IN  MEMORIAM 
FLORIAN  CAJORl 


(X  <3.f^^  I 


CALCULUS  AND  GRAPHS 


THE  MACMILLAN  COMPANY 

NEW  YORK    •    BOSTON   •    CHICAGO    •   DALLAS 
ATLANTA    •    SAN   FRANCISCO 

MACMILLAN  &  CO.,  Limited 

LONDON    •   BOMBAY    •  CALCUTTA 
MELBOURNE 

THE  MACMILLAN  CO.  OF  CANADA.  Ltd. 

TORONTO 


CALCULUS  AND  GRAPHS 


SIMPLIFIED  FOR  A  FIRST  BRIEF  COURSE 


BY 
L.    M.    PASSANO 

ASSOCIATE  PROFESSOR  OF  MATHEMATICS 

IN  THE  MASSACHUSETTS  INSTITUTE  OF  TECHNOLOGY 

AUTHOR   OF    '*  PLANE  AND   SPHERICAL  TRIGONOMETRY  " 


TSitm  fork 

THE  MACMILLAN  COMPANY 

1921 


All  rights  reserved 


^    .  <  '^L^au^'v  I 


Copyright,  1921 
By  the  MACMILLAN  COMPANY 


Set  up  and  electrotyped.     Published  December,  1921. 


Printed  in  the  United  States  of  America 


u> 


PREFACE 

The  importance  of  the  natural  sciences  is  so  generally 
recognized  as  to  need  no  emphatic  statement.  Nor  is  it 
necessary  to  point  out  the  dependence  of  the  sciences  upon 
the  study  and  knowledge  of  mathematics.  This  dependence 
is  closer  and  more  direct  in  the  case  of  calculus  than  in  the 
case  of  any  other  branch  of  mathematics,  unless,  perhaps, 
we  except  elementary  algebra  and  trigonometry.  It  is 
primarily  for  the  purpose  of  making  the  elements  of  the 
calculus  directly  and  familiarly  available  to  students  of 
physics,  chemistry  and  other  sciences  that  the  present  book 
is  written.  At  the  same  time  it  is  hoped  that  the  book  will 
be  found  well  adapted  to  the  use  of  those  who  wish  an  ele- 
mentary knowledge  of  calculus  for  its  cultural  value. 

No  knowledge  of  Analytic  Geometry  is  assumed  on  the 
part  of  students  using  the  present  text.  On  the  other  hand 
the  idea  of  coordinate  axes  and  their  use  in  the  graphical 
representation  and  study  of  simple  algebraic  and  transcen- 
dental functions  is  introduced  in  the  first  chapter  and  used 
continually  throughout  the  work.  The  student  becomes 
familiar  with  the  fundamental  ideas  of  analytic  geometr}'-, 
learns  to  use  both  algebraic  and  geometric  methods  in  the 
study  of  functions,  and  becomes  acquainted  with  the  forms 
and  equations  of  simple  curves  without  definition  of  those 
curves  or  detailed  study  of  their  properties.  The  student 
thus  acquires  all  the  knowledge  of  analytic  geometry  neces- 
sary to  an  understanding  of  the  elements  of  calculus;  and 
assuming  on  his  part  a  knowledge  of  elementary  algebra 
and  trigonometry,  the  calculus  is  made  available  for  a  first 
college  course. 

Such  a  course,  the  writer  believes,  is  not  only  more  useful 


vi  PREFACE 

but  also  more  interesting  and  simple  than  a  first  course  in 
analytic  geometry.  It  is  not  intended,  however,  that  the 
present  work  should  replace  the  study  of  analytic  geometry, 
the  importance  of  which  should  not  be  underrated,  but  that 
the  study  of  the  latter  subject  should  simply  be  deferred 
until  the  more  immediately  important  subject,  the  calculus, 
has  been  acquired. 

The  author  has  striven  to  present  the  subject  in  as  simple  a 
way  as  possible.  The  aim  has  been  to  make  the  student 
understand  the  subject;  not  to  write  a  book  that  would 
satisfy  meticulous  mathematical  pedantry.  In  so  doing  the 
author  may  have  in  places  sacrificed  logical  detail  to  sim- 
plicity of  presentation,  but  never,  he  hopes;  accuracy  of 
statement.  In  the  opinion  of  the  writer  a  too  rigidly  logical 
proof  with  its  paraphernalia  of  subscripted  Greek  letters  is 
out  of  place  in  an  elementary  first  course  in  calculus,  for  the 
reason  that  the  student  never  understands  such  a  proof. 
Or  if  by  arduous  effort  he  does  grasp  its  meaning  it  is  at  the 
expense — in  time  and  labor — of  other  things  that  are  more 
important  and  far  more  useful. 

The  author  wishes  to  thank  a  number  of  his  colleagues  for 
criticism  and  suggestions  in  the  writing  of  the  present  work. 
He  is  especially  indebted  to  Professor  C.  L.  E.  Moore  and 
Professor  D.  P.  Bartlett  for  their  criticism  of  the  manuscript 
and  for  improvements  which  they  suggested. 

L.  M.  Passano. 
Massachusetts  Institute  of  Technology 

Cambridge,  Mass. 


CONTENTS 


ART. 
1. 

2. 
3. 
4. 


Chapter  I.    Variables,  Functions,  and  Graphs 

Variables  and  functions 1 

A  system  of  rectangular  coordinates    .....  3 

Functions  and  graphs;  plotting  graphs         ....  4 
Sketching   graphs;   critical   or  peculiar   points;    zero   and 

infinity;  asymptotes 9 


Chapter  II.  Increments.  Differentiation  and  the  Derivative. 
Theorems  and  Formulas  of  Differentiation  for  Algebraic 
A^D  Trigonometric  Functions 


5.  Increments;  continuous  functions 

6.  The  derivative;  differentiation 

7.  The  derivative  of  a  sum 

8.  The  derivative  of  a  product  . 

9.  The  derivative  of  a  power    . 

10.  Collected  formulas  of  differentiation 


18 
22 
23 
24 
25 
26 


Chapter  III.     Applied  Meanings  and  Uses  of  the  Derivative. 
Rate  of  Change.    Velocity.    Acceleration 


11.  Rate  of  change.    Velocity.    Acceleration 

12.  Angular  velocity  and  acceleration 

13.  Components  of  velocity 

Collected  formulas        .... 


33 
35 
37 
39 


Chapter  IV.  Applied  Meanings  and  Uses  of  the  Derivative, 
Continued.  Direction.  Maxima  and  Minima.  Differen- 
tials, Approximations,  Errors 


14.  Direction  of  a  hne.    Slope    .  ^      . 

15.  Maxima  and  minima    .... 

16.  Differentials.    Approximate  increments. 
Collected  formulas        .... 


Errors  , 


46 
49 
60 
62 


Vll 


viii  CONTENTS 

Chapter   V.      Integration.      Indefinite    Integrals.      Methods 

OF  Integration 
ART.  page 

17.  Integration.    The  indefinite  integral 68 

18.  Methods    of    integration.      Recognition,    transformation, 

substitution 69 

Recognition 69 

Collected  formulas  of  integration 70 

Transformation 70 

Substitution  of  a  new  variable 72 

Chapter  VI.     Integration.    Approximate  Summation.    The  Def- 
inite Integral 

19.  Approximate  summation 80 

Chapter    VII.      Integration  as  a    Summation.      The    Definite 
Integral.    Applications  of  the  Definite  Integral 

20.  Integration  as  a  summation.    The  definite  integral       .         .  88 

21.  General  meaning  of  the  definite  integral       .         .         ,         .  93 

22.  Areas  of  curves 94 

23.  Volumes  of  surfaces  of  revolution 96 

24.  Fluid  pressure       .........  98 

25.  Moment  of  force  or  turning  moment 102 

26.  Distance  determined  from  velocity,  acceleration,  and  time     .  104 

27.  Centre  of  pressure  and  centre  of  gravity       .         .         .         .111 

28.  Mean  value 117 

29.  Work,  attraction,  mass 121 

Chapter  VIII.     Miscellaneous  Examples 

30.  Miscellaneous     examples,     algebraic     and    trigonometric 

functions 126 

Chapter  IX.    Exponential  and  Logarithmic  Functions 

31.  Exponential  functions  and  their  graphs         .         .         .         .132 

32.  Logarithmic  functions  and  their  graphs         .         .         .         .135 

33.  Derivatives  and  integrals  of  exponential  and  logarithmic 

functions 136 

Collected  formulas 139 

34.  Integration  by  parts 140 

35.  The  compound  interest  law  or  the  law  of  organic  growth  145 

Answers 153 


CALCULUS  AND  GRAPHS 


CALCULUS  AND  GRAPHS 

CHAPTER  I 
VARIABLES,  FUNCTIONS,  AND  GRAPHS 

1.  Variables  and  Functions.  In  dealing  with  numbers 
one  is  brought  to  consider  two  well  marked  kinds :  constants 
and  variables.  Their  names  suggest  their  nature.  A 
constant  number,  or  quantity  is  one  which,  under  the  con- 
ditions of  the  problem  in  which  it  occurs,  does  not  change 
in  value;  a  variable  is  one  which  takes  different  values. 
Whether  the  values  of  either  are  known  or  unknown  makes 
no  difference.  Of  two  numbers,  a  and  x,  say,  we  may  not 
know  the  single  actual  value  of  a,  and  may  know  that  the 
X  must  have  the  definite  range  of  values  from  2  to  6.  The 
number  a  would  still  be  a  constant  and  the  number  x  a 
variable. 

In  many,  or  most,  of  our  problems  it  happens  also  that 
we  have  to  deal  with  two  variables,  and  any  number  of 
known  or  unknown  constants;  and  with  two  variables  so 
related  that  the  value  of  one  depends  upon  the  value  of 
the  other.  For  example,  the  room  in  which  we  are  sitting, 
the  doors  and  windows  being  shut,  will  have  a  certain 
temperature  which  varies.  We  can  change  the  tempera- 
ture by  increasing  or  decreasing  the  amount  of  steam  in 
the  radiator.  The  amount  of  steam  we  can  regulate  in  any 
way  we  please,  but  the  temperature  we  can  regulate  only 
by  means  of  the  steam,  upon  which  it  depends.  We  ex- 
press this  mathematically  by  saying  that  the  temperature, 
a  dependent  variable,  is  a  function  of  the  steam,  an  inde- 

1 


2  CALCULUS  AND  GRAPHS 

pendent  variable,  and  we  write  the  relation  thus,  using  y 
to  represent  temperature  and  x  steam, 

y  =  f  W,  ory  =  F  {x)    y  ==  ^  {x)  etc.; 

read,  y  equals  f  oi  x,  ^  of  a:,  etc.,  or,  in  general,  y  equals  a 
function  of  x. 

Many  simple  examples  of  functions  suggest  themselves. 
For  example,  A  =  S^  expresses  the  fact  that  the  area  of  a 
square  {A)  is  a  particular  function  of  its  side  (aS);  c  =  27rr, 
expresses  that  the  circumference  of  a  circle  (c)  is  a  particular 
function  of  its  radius  (r).  By  giving  values  to  s  in  the  one 
case  or  to  r  in  the  other  one  can  find  the  corresponding 
values  of  A  or  c.  Thus,  when  the  side  of  the  square  is  2 
feet  the  area  is  4  square  feet;  when  s  is  12  feet,  A  =  144 
sq.  ft.;  when  r  is  7  inches  c  is  Mtt  =  44  inches. 

EXAMPLES 

1.  Express  the  volume  of  a  cube  as  a  function  of  the  edge.  Find 
the  value  of  the  volume  when  the  edge  is  2  feet,  7  feet,  3.1  centimeters. 

2.  Express  the  volume  and  surface  of  a  sphere  each  as  a  function 
of  the  radius;  the  volume  as  a  function  of  the  surface. 

3.  The  height  of  a  rectangular  pyramid  is  three  times  the  side  of 
its  base.  Express  the  volume  of  the  pyramid  as  a  function  of  the  side 
of  the  base;  as  a  function  of  the  height. 

4.  The  illumination,  /,  from  a  source  of  light  varies  directly  as  the 
candle  power,  c,  and  inversely  as  the  square  of  the  distance,  d.  If 
the  distance  is  10  feet  express  the  illumination  as  a  function  of  the 
candle  power.  If  the  candle  power  is  50,  express  the  illumination  as 
a  function  of  the  distance.  Which  gives  the  greater  illumination  .5 
candle  power  at  10  feet  or  50  candle  power  at  100  feet? 

Find  the  values  of  the  following  functions  when  x  =  1  and  when 
a;  =  4;  also  the  amount  by  which  the  function  increases  or  decreases 
as  X  changes  in  value  from  1  to  4. 

5.  2/  =  ^^  +  2  a:  10.  y  =  2^ 

6.  s  =  3\/^'  11.  2/  =  a;  — 3a;2 


VARIABLES,   FUNCTIONS,  AND  GRAPHS 


7,  u  =  2x  + 

S.  y  =  logiox. 
9.  y  =  tan  x 


2x 


12.  y 


13.  y  =—'5x  +  - 
x 


14.  y 


■  x^  -}-  x 


2.  A  System  of  Rectangular  Coordinates.  Let  us  draw 
in  a  plane  two  straight  lines  at  right  angles  to  each  other, 
(Fig.  1). 


■4 

--3-- 


R 


X 


Q 


^ — h 


-4   -3    -2  -1  0 


1 


-f- 


P 


H h 


12     3     4     6 


•  -1 

-2 
•■-3 
•-4 

F 

Fig.  1 

Let  US  call  the  point,  0,  in  which  the  lines  intersect,  the 
origin,  the  lines  themselves  the  axes,  of  the  system  of 
rectangular  coordinates.  To  distinguish  the  two  axes  one, 
the  line  XOX  is  called  the  axis  of  x  or  the  axis  of  abscissas; 
the  other  70  F,  the  axis  of  y  or  the  axis  of  ordinates. 
Suppose  we  wish  to  distinguish  the  points  on  OX  from 
each  other.  To  do  so  we  associate  with  each  point  a  sym- 
bol, called  a  number.  This  might  be  done  in  any  way, 
but  for  the  sake  of  system  and  order  we  will  lay  off  any 
convenient  unit  of  length  along  XOX  and  attach  to  the 
points  thus  got  to  the  right  of  0  the  series  of  numbers 

1,  2,  3, 4 ;  to  those  to  the  left  of  0  the  series  —  1,  — 2, 

—  3,  —  4,  Thus,  using  fractional  and  incom- 


4  CALCULUS  AND  GRAPHS 

mensurable  numbers  as  well  as  integers,  there  corresponds 
to  every  point  on  XOX  a  certain  number  and  to  every 
number  a  point  on  XOX.  In  the  same  way  to  every  point 
on  YO  Y  corresponds  a  number,  and  to  every  number  a 
point  on  YO  F.  But  given  the  number  2  we  should  not 
know  whether  the  point  P  were  meant  or  the  point  E 
(Fig.  1).  To  remove  this  ambiguity  we  say,  when  the 
point  P  is  meant,  the  point  whose  abscissa,  or  a;,  is  2;  when 
R  is  meant,  the  point  whose  ordinate,  or  y,  is  2;  written 
respectively  x  =  2  and  y  =  2, 

Suppose  next  that  we  wish  to  consider  a  point  >S,  not 
lying  on  either  axis.  Obviously  its  abscissa,  distance  along 
or  parallel  to  OX  is  4  and  its  ordinate,  distance  along  or 
parallel  to  OF  is  3,  which  we  write  a:  =  4,  t/  =  3  *,  or  more 
briefly  (4,  3).  Thus  to  every  point  in  the  plane  corresponds 
a  pair  of  numbers  and  to  every  pair  of  numbers  a  point  in 
the  plane.  It  will  be  noticed  also  that  the  system  of  co- 
ordinates is  a  connecting  link  between  the  simple  geo- 
metrical concept,  a  point,  and  the  simple  algebraic  concept, 
a  number,  and  thus  in  general  between  the  two  branches 
of  mathematics;  so  that  each  may  be  used  as  a  help  in 
studying  the  other. 

3.  Functions  and  Graphs:  Plotting  Graphs.  Consider  y 
a  simple  function  of  x;  say,  y  =  2  .t  +  3.  We  can  give  the 
independent  variable  x  any  values  we  please,  and  compute 
the  corresponding  values  of  the  function  y,  as  shown  in  the 
following  table: 


X 

y 

0 
3 

1 

5 

2 

7 

3 

9 

—  1 

—  2 

—  3 

etc. 

1 

—  1 

—  3 

etc. 

*  Obviously  any  other  letters  might  be  used.    Thus  OT  might  be  the  axis  of  abscissas 
and  OS  the  axis  of  ordinates.    The  point  would  then  be  written  f  =  4,  s  =»  3. 


VARIABLES,  FUNCTIONS,  AND  GRAPHS 


We  thus  have  seven  pairs  of  numbers,  values  of  x  and  2/, 
and,  as  we  have  seen,  each  such  pair  of  numbers  corresponds 
to  a  point  in  a  system  of  coordinates.  If  we  mark  these 
points,  plot  them  as  it  is  called,  we  shall  have  a  pic- 
ture, a  graphic  representation,  of  the  functional  relation 
connecting  the  two  variable  numbers  x  and  y  as  shown 
in  Fig.  2. 

If  we  take  values  of  a:  ^ 

at  small  intervals  we  can 
get  as  many  points  as  we 
please  as  near  together 
as  we  please,  the  graph 
appearing  more  and  more 
nearly  as  a  continuous  ]L 
line.  We  thus  speak  of  y 
as  a  continuous  function 
of  x;  a  concept  that  will 
be  more  strictly  defined 
later  on.  (See  Art.  5.) 
We  can  now  study  2/,  the 
function  of  x,  by  using 
the  graph  and  the  methods  of  Geometry  or  by  using  the 
equation  and  the  methods  of  Algebra.  In  many  cases, 
indeed,  we  use  both.  Note  most  carefully  that  if  a  point 
lies  on  the  graph  its  coordinates  must  satisfy  the  equation 
of  the  graph;  if  a  pair  of  values  of  the  coordinates  satisfy 
the  equation  the  corresponding  point  must  lie  on  the  graph 
of  the  function.  Also  if  a  point  lies  at  the  same  time  on 
two  (or  more)  graphs,  its  coordinates  will  satisfy  the  two 
(or  more)  equations  of  the  graphs.  If  we  wish  to  find 
values  of  the  variables  {x  and  y^  say)  that  satisfy  two 
equations  we  may  sketch  the  graphs  of  the  equations  and 
measure  the  abscissas  and  ordinates  of  their  points  of  inter- 
section; or,  if  we  wish  to  find  the  points  of  intersection  of 
two  graphs  we  can  solve  simultaneously  the  equations  of 


FiQ.  2 


6  CALCULUS  AND  GRAPHS 

the  graphs.    Thus  the  graphs  of  y  =  2  x  —  3  and  x  +  y  =  4 
intersect  in  the  point  x  =  ^,  y  =  ^, 
Let  us  consider  other  simple  functions. 

Example  L     y  =  2x^  —  S,     See  Fig.  3. 


X 

y 

0 

1 

2 

3 

—  1 

—  2 

—  3 

etc. 
etc. 

—  8 

—  6 

0 

10 

—  6 

0 

10 

Example  2.     y  =  sin  x.     See  Fig.  4. 

In  sketching  this  graph  it  is  convenient,  since  x  must  be 
measured  along  OX,  to  express  the  angle  in  circular 
measure. 


X 

y 

0 
0 

X 

6 
.50 

X 

3 

X 

2 

2x 
3 

5  X 
6 

X 

7x 
6 

4  X 
3 

3x 
2 

5x 
3 

11  X 

6 

2x 

.87 

1.00 

.87 

.50 

0 

—  .50 

—  .87 

—  1.00 

—  .87 

—  .50 

0 

VARIABLES,  FUNCTIONS,  AND  GRAPHS        7 

We  need  not  give  values  to  x  greater  than  27r  because  the 
trigonometric  functions  are  periodic;  that  is,  the  same 
value  of  the  function  will  occur  at  regular  intervals  of  27r. 
Note  that,  as  a  matter  of  convenience,  different  units  are 
used  on  the  two  axes. 


FiQ.  4 


The  graph  extends  indefinitely  to  right  and  left  in  repeti- 
tion of  the  part  drawn,  as  is  indicated  by  the  dotted  por- 
tion in  Fig.  4. 

Y 


Example  3.  y  =  2  cos  3  x.     See  Fig.  5. 


CALCULUS  AND  GRAPHS 


X 

0 

X 

18 

X 

9 

X 

6 

2x 
9 

5x 
18 

X 

3 

7x 
18 

4  X 
"9~ 

X 

2 

X 

18 

X 

~"  9 

X 

"~"  6 

^x 

0 

X 

6 

X 

3 

X 

2 

2x 
3 

5x 
6 

X 

7x 
6 

4  X 
~3" 

3  X 
2 

X 

~^  6 

X 

~"  3 

X 

~^  2 

y 

2 

1.74 

1 

0 

—  1 

—  1.74 

—  2 

—  1.74 

—  1 

0 

1.74 

1 

0 

Note  carefully  in  this  example  that,  while  we  use  values  of 
3  a:  to  compute  values  of  y^  we  plot  the  points  by  using  values 
of  X  with  values  of  y.  Note  also  that  for  convenience  we 
use  a  smaller  unit  along  0  Y  than  along  0  X, 


EXAMPLES 

1.  What  are  the  coordinates  of  the  origin?  What  is  the  abscissa  of 
any  point  on  the  axis  of  yf    The  ordinate  of  any  point  on  the  axis  of  xf 

2.  Do  the  points  (1,  3)  (1,  4)  (0,  —  1)  (3,  —2)  lie  on  the  graph  of 
the  function  y  =  5x  —  1?  Do  they  lie  on  the  graph  oi  y  =  2  x^  -\-  1? 
Why? 

3.  What  points  are  represented  by  the  equation  a:  =  3?  By  the 
equation  2/  =  —  2?    By  a:  =  0?    By  y  =  0? 

Plot  the  graphs  of  the  following  functions: 


4.  2  2/  =  X  +4 

5.  y  =  S  x 


6.^=4-3 

7.  y  =  (x  — 1)2 


y  = 


9.  y  =  X  —  x^ 

X 

10.  y  =  sin;^ 

11.  2/  =  3  cos  2  a; 

12.  y  =  tan  x 

13.  2/  =  sec  X 


14.  The  distance,  s,  through  which  a  body  falls  in  a  time,  t,  is  ex- 
pressed by  the  equation  s  =  16  t^.  Express  graphically  the  distance  as 
a  function  of  the  time. 

15.  Express  graphically  the  volume  of  a  cube  as  a  function  of  the 
edge. 


VARIABLES,  FUNCTIONS,  AND  GRAPHS        9 

16.  Draw  a  graph  showing  the  area  of  a  circle  as  a  function  of  the 
radius. 

17.  A  particle  moves  so  that  its  distance  from  the  axis  of  y  is  given 
by  X  =  2  /,  its  distance  from  the  axis  oi  xhy  y  =  t  —  2,  where  t  repre- 
sents the  time,  in  seconds,  the  particle  has  moved.  Plot  the  path  of 
the  body  during  the  first  10  seconds.  From  what  point  does  the  particle 
start?    At  what  point  is  it  at  the  end  of  the  tenth  second? 

Find  the  points  of  intersection  of  the  following  pairs  of  graphs: 


18. 

y  =  2a; --3 

19. 

2x+Sy  =  7 

20. 

X  +2y  =2 

X  =-y 

X  —  2/  =  4 

Sx  —  ^y  =  1 

21. 

y  =  X 

22. 

y  =  x  +  l 

23. 

y  =  X 

9  ^  16 

x2  H-  2/2  =  16 

y  =  4:X^ 

24. 

2/=2 

25. 

y  =  1 

26. 

y  =  a/3 

y  =  3  sin  X 

X 

2/  =  cos  2 

y  =  tan  x 

27.  Two  particles  move  in  straight  lines,  passing  through  an  origin  0, 
according  to  the  laws  s  =  Zt  —  5  and  s  =  ^  +  4  respectively.  At 
what  time  (0  will  the  two  particles  be  at  equal  distances  (s)  from  the 
origin?    What  is  the  distance? 

Two  particles  move  in  a  straight  line  passing  through  an  origin  0, 
according  to  the  laws  s  =  ^  +  4  and  8  =  1"^  —  5. 

28.  When  will  the  two  particles  be  at  equal  distances  from  0?  What 
is  the  distance? 

29.  When  will  the  first  particle  be  twice  as  far  from  0  as  the  second? 
What  is  the  distance  of  each? 

30.  When  will  the  second  particle  be  twice  as  far  from  0  as  the  first? 
What  is  the  distance  of  each? 

4.  Sketching  Graphs:  Critical  or  Peculiar  Points.  In 
many  of  our  problems  it  happens  that  we  are  interested 
not  so  much  in  the  exact  figure  of  the  graph  representing 
a  function  as  in  its  general  features;  whether,  for  instance, 
it  is  a  closed  or  open  curve;  what  are  its  limits;  the  position 
of  critical  or  peculiar  points.    Such  information  can  usually 


10  CALCULUS  AND  GRAPHS 

be  got,  without  accurately  plotting  the  graph,  by  sketching 
it  with  especial  reference  to  its  peculiarities.  The  following 
examples  illustrate  some  of  the  methods  employed  for  the 
ends  mentioned. 

Example  L         y  +  2  =  3  (x—iy. 

Fut    y  +  2  =  y',       x  —  1  =  x^ 

The  equation  then  becomes  ^'  =  3  x^^. 
It  is  obvious  that  numerically  equal  positive  and  negative 
values  of  x^  give  equal  values  of  ?/',  so  that  the  curve  is 
symmetrical  with  respect  to  0Y\*  Also,  x^  occurring  as 
a  square  only,  y'  can  not  be  negative.  The  least  value  of 
2/',  zero,  is  got  when  x^  is  numerically  least  (=0).  The 
curve  can,  therefore,  be  sketched  readily  without  the  labor 
of  obtaining  numerical  points.    Fig.  6. 

This  is  not,  however, 
the  graph  of  the  function 
originally  given  but  of  the 
simpHfied  function  y^  = 
3  x^^.    But,  by  the  assump- 

^i  tions  we  see  that  y  —  y' — 

2  and  x  =x'  +  1;  that  is, 
the  distance  {y)  from  the 
true  axis  of  a;  is  2  less  than 
the  distance  from  OX'  and 
the  distance  {x)  from  the 
true  axis  of  2/  is  1  more  than  the  distance  from  0Y\  We 
therefore  draw  OX  parallel  to  and  2  units  above  OX' 
and  OY  parallel  to  and  1  unit  to  the  left  of  OY',  The 
axis  of  symmetry  of  the  curve  is  now  seen  to  be  a:  =  1, 
and  the  vertex  of  the  curve  x  =  1^  y  =  —  2,  as  shown  in 
Fig.  7. 

*  In  general  it  is  worth  while  to  note  that  if  the  equation  contains  even  powers  only 
of  xiy)  it  is  symmetrical  with  respect  to  OYiOX). 


o' 


VARIABLES,  FUNCTIONS,  AND  GRAPHS      11 


Exam-pie  2,    y 


=  sin  (^x  +  y . 


TT 


Put  X  +  J-  =  ^'  SO  that  7j  =  sin  x\ 

We  can  simplify 
the  sketching  of  y 
=  sin  x^  by  using 
our  knowledge  of 
the  trigonometric 
functions.  The 
greatest  value  of  ^ 
the  sine  of  an  an- 
gle is  1;  the  least 
value,  —  1;  the 
least  numerical 
value,  0.  There- 
fore, put  sin  x^  =  0. 
ar'=  sin"^  0,  and 
we  have 


x' 

0 

IT 

27r 

etc. 

y 

0 

0 

0 

etc. 

the  points  where  the  graph  crosses  0  X\ 

Put  sin  x'=  1,0:'=  sin"^  1   Put  sin  x'  =  —  1,  a:'  =  sin"^( —  1) 


TT 

5  TT 

Stt 

7  TT 

x' 

~2 

2 

etc. 

x' 

2 

2 

etc. 

y 

1 

1 

etc. 

y 

—  1 

—  1 

etc. 

the  highest  points  on  the 
graph. 


the  lowest  points  on  the 
graph. 


12 


CALCULUS  AND  GRAPHS 


We  can  now  sketch  the  graph  by  means  of  these  critical 
points  and  fix  the  original  axes  as  in  Example  1  above. 
See  Fig.  8  and  compare  Example  2,  Art.  3. 


Fig.  8 

Example  3.     y  =  x^  +  2  x'^  =  x^x  +  2) 
Put  0^2  =  0        and         a:  +  2  =  0 
ora:  =  0  x  =  —  2 

thus  obtaining  critical  points.  Then  by  taking  values  of  x 
less  than  and  greater  than  the  critical  values  proceed,  as 
shown  in  the  following  table,  to  determine  the  peculiarities 
of  the  graph  and  to  sketch  it.    Fig.  9. 


X 

y 

—  3 

—  2 
0 

—  1 

+ 

0 
0 

1 

+ 

The  only  points  actually  found  are  ( — 2,  0)  and  (0,  0). 
Since  y  is  negative  to  the  left  of  x  =  —  2  the  curve  comes 
up  from  below  the  axis  of  a:,  which  it  crosses  at  a:  =  —  2; 
remains  above  OX  between  x  =  —  2  and  x  =  0,  since  in 
this  interval  y  is  positive;  touches  the  axis  oi  x  Sit  x  =  0, 


VARIABLES,   FUNCTIONS,  AND  GRAPHS      13 


and  extends  above  OX  thenceforth  because  y  remains 
positive  for  values  of  x  greater  than  x  =  0.  It  must  be 
noted  that  the  graph  is 
not  accurately  plotted  but  ^ 

merely  sketched.  We  do 
not  know,  for  instance,  the 
highest  point  on  the  arch 
of  the  curve;  that  is,  the 
greatest  value  of  the  func- 
tion lying  between  the 
values  when  x  =  —  2  and 
a:  =  0.  Later  in  our  work 
we  shall  learn  how  to  deter- 
mine such  values. 


Fig.  9 


Example  4-     ^y  —  2x-\-2y-\-S  =  0. 

In  this  case  y  may  be  regarded  as  a  function  of  a:  or  a: 
as  a  function  of  y.  Solving  the  equation  for  y  and  x  re- 
spectively we  have 


y  = 


2x 


x+2 


X  = 


S  +  2y 
2-y  ' 


Using  the  former  and  putting 
3  =  0 


2x 

X 


_    3 
—    IS 


0:+  2  =  0 

X  =  —2 


we  have  critical  points.  As  in  Example  3,  a:  =  f  gives 
2/  =  0,  but  X  =  —  2  gives  y  =  —  ^.  What  does  this  mean? 
Zero  does  not  mean  "  nothing."  Mathematics  does  not 
concern  itself  with  things  that  mean  nothing.  On  the  con- 
trary the  science  prides  itself  on  the  fact  that  every  concept 
and  symbol  used  has  a  very  definite  and  exact  meaning,  so 
that  once  the  *^  language  "  is  mastered  there  can  be  no 
doubt  as  to  the  meaning  of  every  word  and  sentence. 


14 


CALCULUS  AND  GRAPHS 


The  symbols  -  or,  in  general,  -,  mean  simply  that  a 

(or  7)  is  being  divided  by  a  number  which  grows  smaller 
and  smaller  *  and  which  can  be  made  as  small  as  we  please. 
As  the  divisor  becomes  smaller  and  smaller  the  value  of 
the  quotient,  or  fraction,  becomes  greater  and  greater;  and 
by  taking  the  divisor  small  enough  we  can  make  the  value 
of  the  fraction  as  great  as  we  please.  We  may  express 
this  by  saying  that,  as  the  denominator  approaches  zero 
as  a  Kmit,  the  fraction  increases  without  limit,  and  may 


write  it,  briefly,  t:  =  ^ 


Thus  in  the  above  example,  as 


X  becomes  more  and  more  nearly  equal  to  —  2,  a;  +  2  be- 
comes smaller  and  smaller,  and  by  taking  x  as  near  —  2 
as  we  please  (on  either  side)  we  can  make  a:  +  2  as  small 
as  we  please.  We  express  this  briefly  by  saying  that  when 
X  equals  minus  two,  x  plus  2  equals  zero  and  the  fraction 
equals  infinity,  and  write 


X  —  —  2,  a^+2  =  0,  and  y  = 
We  do  not  know,  however,  whether  y  is  plus  infinity  or 


7 


*  Numerically;  that  is,  without  regard  to  the  algebraic  sign. 


VARIABLES,   FUNCTIONS,  AND  GRAPHS      15 

minus  infinity,  since  we  do  not  yet  know  whether  x  +  2 
is  plus  zero  or  minus  zero;  that  is,  whether  x  -{-  2  is  an 
exceedingly  small  positive  or  an  exceedingly  small  negative 
number.  We  will  proceed  to  sketch  our  graph,  however, 
by  marking  (Fig.  10)  the  point  (|,  0)  where  it  crosses  OX, 
and,  where  x  =  —  2,  by  drawing  a  line  parallel  to  OY. 
We  will  also  tabulate  our  values  in  order,  finding  values, 
or  the  sign,  of  7j  for  values  of  x  less  than  and  greater  than 
the  critical  values. 


x 

—  3 

—  2 

0 

3 

2 

y 

+  9 

CX) 

3 
2 

0 

1 

T 

Since  y  is  positive  to  the  left  oi  x  =  —  2  we  see  that  the 
ordinate  is  positive  and  very  large  just  to  the  left  of  a;  = 
—  2;  expressed,  ?/  =  +  oo .  Since  y  is  negative  between 
X  =  —  2  and  a:  =  f  we  see  that  the  ordinate  is  negative 
and  very  large  just  to  the  right  of  x  =  — 2;  expressed, 
y  =  —  00 .  We  might  now  proceed  to  sketch  the  graph, 
but  before  doing  so  let  us  consider  the  other  form  of  the 
equation, 

3  +  2t/ 


X  = 


y 


and  we  find  x  =  0,  a 


Put        3  +  22/ =  0,      2/  =  — I 

point  already  discussed. 

Put        2  —  y  =  Oj  y  =  2        and  we  find  a:  =  oo  .     We, 

therefore,  draw  a  line  {y  =  2)  parallel  to  0  X,  and  tabulate 

our  values. 


y 

—  2 

3 

0 

2 

3 

X 

1 

T 

0 

3 

00 

—  9 

16  CALCULUS  AND  GRAPHS 

Reasoning  as  above  we  find  x  =  —  oo  for  positions  just 
above  the  line  y  =  2,  and  a;  =  +  oo  just  below  the  hne 
y  =  2.  The  graph  can  now  be  completely  sketched  as 
shown.  The  lines  x  =  —2  and  ?/  =  2  are  called  asymptotes 
of  the  curve.  They  meet  the  curve  at  a  distance  greater 
than  any  number  one  may  please  to  assign  however  great; 
that  is,  at  infinity. 

EXAMPLES 

Sketch  the  graphs  of  the  following  functions: 

1.  2/  — 3  =  (a;  — 1)2  b.  y  =x^  -\-x 

2.  y +4  =  2  (2a;  — 3)2  6.  i/  =  x^  ^  x 

3.  X  =  (y  —  2)2  7.  X  =y^^2  y^ 

4.  X  =  (y  +  3)2  S.  X  =  y^  -\-3  y^ 

9.  2/  =  sin  (x  —  |)  13.  y  =  cos  {x  +  1) 


10.  y  =  sm{x  +2  %) 

14.  y  =  cos  (x  —  1.) 

11.  2/  +  2  =  2sinaj 

15.  y  =  tan  (x  —  ^ 

12.  2/  —  3  =  sin  2  a; 

16.  y  =  tan  (x  + 1) 

17.  xy  —  X  -{-2y  =  1  21.  xy  —  x  —  by  +  a  =  0 

IS.  xy^x  —  2y  +  1  =  0  22.  xy  +  x  +  by  =  a 

19.  2  x?/  —  a:  +  6  ?/  +  2  =  0  23.  s  =  ^^  _  ^2^ 

20.  2xy  +x+6y  =  2  24.  s  =  t^  -{- bt^ 

25.  The  product  of  the  pressure  (p)  and  volume  (v)  of  a  perfect 
gas  is  constant  and  equal  in  a  particular  case,  to  4.  Express  the  volume 
as  a  function  of  the  pressure  and  sketch  the  graph.  What  is  the  volume 
when  the  pressure  equals  1000?  What  happens  when  the  pressure  in- 
creases without  limit?  Does  v  ever  become  zero?  Does  the  volume 
ever  vanish? 

26.  A  particle  moves  in  a  straight  line  so  that  s  =  t^  —  3  ^2^  where  s 
is  the  distance  in  feet  from  an  origin  on  the  line  and  t  is  the  time  in 
seconds.    Sketch  the  graph  of  the  function.    By  means  of  your  graph 


VARIABLES,  FUNCTIONS,  AND  GRAPHS      17 

determine  at  what  time  the  particle  will  be  at  the  origin  on  the  straight 
line.    Does  it  move  to  the  left  or  to  the  right  of  the  origin?    How  long? 

27.  A  particle  moves  in  a  straight  line  so  that  s  =  t^  —  3^  +  2, 
s  and  t  being  distance  and  time  as  in  Example  26.  Sketch  the  graph 
of  the  function.  By  means  of  your  graph  determine  from  what  point 
on  the  straight  line  the  particle  starts  to  move.  Does  the  particle  ever 
reach  the  origin  on  the  straight  line?  When?  Does  it  move  to  the 
left  or  to  the  right  of  the  origin?    How  long? 

28.  A  particle  moves  so  that  its  distance  from  OY  is  always  x  =  t; 
its  distance  from  OX  is  ij  =  P  -]-  2  t,  where  t  is  the  time  in  seconds. 
Sketch  the  path  of  the  particle.  Where  does  the  particle  start  to  move? 
Does  it  ever  move  backward  in  its  path?    Why? 


CHAPTER  II 

INCREMENTS.  DIFFERENTIATION  AND  THE  DE- 
RIVATIVE. THEOREMS  AND  FORMULAS  OF 
DIFFERENTIATION  FOR  ALGEBRAIC  AND  TRIGO- 
NOMETRIC FUNCTIONS 

6.  Increments:  Continuous  Functions.  In  many  prob- 
lems it  is  of  greater  interest  and  importance  to  know  how 
and  by  how  much  a  variable  is  changing  than  to  know  the 
actual  value  of  the  variable  itself.  In  the  case  of  the  inde- 
pendent variable  this  is  not  so,  because  by  the  very  nature 
of  an  independent  variable  we  can  make  the  change  any- 
thing we  please.  But  if  we  make  a  change  in  the  value  of 
an  independent  variable,  x,  what  change  is  caused  in  the 
value  of  a  variable,  y,  which  depends  upon  x;  in  the  function 
of  X  which  y  represents?  Let  us  consider  a  simple  case, 
y  =  x^.  Suppose  x  has  the  value  2  (or  4)  and  that  we  make 
changes  in  its  value,  give  it  various  increments,  as  they  are 
called,  w^here  the  increment  may  be  either  positive  or  nega- 
tive. The  resulting  values  and  changes  in  value  of  y  are 
shown  in  the  following  table, jv^rejbhe  symbol  Ax  (delta  y) 
means  the   increment  given   to  ^  JjSJ:^4^M  the  increment 


caused  in  ?y  bv  A  : 

r. 

X 

y 

^x 

A2/ 

^y 

Ax 

X 

y 

Aa; 

Ay 

Ay 

A~x 

2 

4 

4 

16 

3 

9 

1 

5 

5 

5 

25 

1 

9 

9 

2.1 

4.41 

.1 

.41 

4.1 

4.1 

16.81 

.1 

.81 

8.1 

2.01 

4.0401 

.01 

.0401 

4.01 

4.01 

16.0801 

.01 

.0801 

8.01 

18 


INCREMENT  AND  DERIVATIVE 


19 


The  tables  show  clearly  that  the  increment  of  the  func- 
tion (  A  I/)  depends  both  upon  the  value  of  the  independent 
variable  (x)  and  its  increment  (Ax);  and  also  show  that 
A  y  grows  smaller  as  A  x  does.  It  would  seem  from  the 
table  that  by  making  A  x  small  enough  we  could  make 
A  2/  as  small  as  we  please.  When  this  is  the  case,  when 
the  increment  of  the  function  can  be  made  smaller  than 
any  number  we  may  assign,  however  small,  by  making  the 
increment  of  the  independent  variable  small  enough — as 
it  is  usually  expressed:  if  the  increment  of  the  function  ap- 
proaches zero  as  a  limit  when  the  increment  of  the  inde- 
pendent variable  approaches  zero  as  a  limit;  written 
limit    A  y  =  0  —  in  such  case  y  is  called  a  continuous 

function  of  :c.* 

The  last  column  of  the  tables  gives  the  value  of  the  ratio 
of  the  increment  of  the  function  to  the  increment  of  the 
independent  variable,  which  ratio  also  is  seen  to  depend 
upon  both  X  and    A  x.     We  note  also  that,  when  x  ^  2, 


the  value  of 


Ay 
A  X 


as   A  a:  grows  smaller  becomes  more  and 

more  nearly  equal  to  4;  when  a:  =  4,  the  value  of  — r-^ 

becomes  more  nearly  equal  to  8  as  Ax  grows  smaller.    We 

might  ask  what  will  be  the  value  of   A  y  and  more  par- 

A  y 
ticularly  of  —-^  when    A  x  has  become  smaller  than  any 


"That  limit      ^y  ia  not  necessarily  zero  may  ba  indicated  by  the  simple  functi(»n 

Ax=  0 


1 
y  =  - 


As  X  grows  nearer  to  —  .1  (  Ax 
decreases)  Av  increases  rapidly. 


X 

y 

Ax 

Ai/ 

—.1 

—10 

.1 

+10 

.2 

20 

.01 

100 

.11 

no 

.001 

1000 

.101 

1010 

.0001 

10000 

.1001 

10010 

20  CALCULUS  AND  GRAPHS 

number  we  may  please  to  assign.  That  is  are  we  justified 
in  saying  that  for  the  function  y  =  x^  when 

x  =  2,     limit      ^  =  4,  and 

Ax=0  ^  X 

X  =  4,     limit      ^  =  8? 

Aa;=0  ^  X 

To  answer  this  question  let  us  try  to  generahze  our 
process.  Let  A  a;  =  /i,  any  number.  Then  the  new  value 
of  the  function  will  he  y  +  /^  y  =  (2  +  h)^  and 

Ay  =  (y  +  Ay)-y  =  (2  +  hf  -  {2f 

which  may  be  written 

and 

A  X 
Therefore,  passing  to  the  limit,  since  h  =  L  x^ 
limit       ^y  _  limit    (a  a.  ■h\  —  a 

Ax  =  0  -^  ~      h=0^'*'"^>'~^ 

Thus,  we  see,  our  conclusion  was  correct;  we  ^'  guessed 
right.''  In  the  same  way  we  could  show  that  for  a:  =  4 
the  result  would  be  8,  and  by  the  same  process  we  could 
find  the  result  for  any  other  numerical  value  of  x.  But 
let  us  generalize  again  and  use  any  value  of  x  whatever. 
Thus 

y  =  x'^^y  +  i^y  =  {x+  hY^  and 

Ay  =  {x+hf  —  x''  =  2hx  +  h^ 

-^  =  2x  +  h, 

A  X 

limit      Al  =  limit    (2x  +  h)  =  2x. 


INCREMENT  AND  DERIVATIVE  21 

Now,  by  simply  putting  any  numerical  value  for  x  in  our 
result,  we  can  find  the  value  of  l™it      -—^  for  the  func- 

Ax  =  0    A  X 

tion  y  =  x^  for  any  value  of  x  whatever. 

The  above  process  is  fundamentally  important  and  can 
be  applied  to  any  of  the  functions  with  which  we  deal.  It 
must  be  thoroughly  mastered. 

Example  1.    Let  y  =  -  and  A  x  =  h,  then 
x 

.     .  1 


Ay  = 

Ajy  ^ 
A  X 

1 
/ix  =  o  IX  X         h  =  o  \     x"-  -f-  nx/  x^ 

which  when  .r  =  1,  2,  3,  etc.,  takes  the  values  —  1,  —  j, 
-jj,  et/C. 

Example  2.     y  =  x^  +  Xy  A  x  =  h, 

y  +  Ay  =  (x+hr^+(x  +  h) 

Ay  =  (x  +  hY  +  {x  +  h)—x''  —  x  = 
2xh  +  hr-\'h 

4^  =  2x  +  A+l 

A  X 

Umit      Al  =  2  .T  +  1. 
Ax  =  o   A  X 

EXAMPLES 

In  the  following  functions  find  the  limit  of  the  ratio  of  the  increment 
of  the  function  to  the  increment  of  the  independent  variable  as  the 
increments  approach  zero  as  a  limit,  showing  each  step  of  the  process. 


1        1 

x-\-  h       x~  x'- 
1 

~h 
+  hx 

x^-\-hx 
limit,  (           ^ 

_^_ 

22  CALCULUS  AND  GRAPHS 


1.  y  =2x^  —  3  2.y  =  -+x  3.  y  =3x^  —  2  x 

4 

4.    s   =   ^3   _|_  1  5     -j^   =  ^3  _|_  _  (^     s    =  ^2  _  _ 

V  f2 

7.  a:?/  =  a:  +  1  8.  x^y  =  o;^  +  x'"^  9.  y  =  V  a; 

^    _L    J  J  X 

10.  2/  = 11.  ?/  =  r— —  12.  pi;  =  a 

^  —  2  3  +  X 

6.  The    Derivative:    Differentiation.      The    expression, 

limit    -—J.    found  in  Art.  5  is  called  a  derivative;  the 

process  of  finding  a  derivative  is  called  differentiation. 
Thus,  y  being  a  continuous  function  ofx  the  derivative  of  y  with 
respect  to  x  is  the  limit  of  the  ratio  of  the  increment  of  y  to  the 
increment  of  x  as  the  increments  approach  zero  as  a  limit. 

The  notation  used  in  Art.  5  is  exact  but  awkward.  The 
student  having  now  learned  the  exact  meaning  of  derivative 
we  may  use  a  simpler  notation,  thus: 

limit      Al  =  ^  (1) 

Ax  =  o   /^x       dx 

dv 
where  dy  and  dx  are  not  used  separately,  and  ~  (read,  the 

ax 

derivative  of  y  with  respect  to  x;  or,  dy  over  dx)  is  not  as 
yet  to  be  treated  as  a  fraction,  though  we  shall  treat  it  so 
later  and  in  many  cases  use  dy  and  dx  advantageously  as 
separate  quantities. 

The  fundamental  process  (Art.  5)  being,  however,  too 
cumbersome  for  frequent  use  we  proceed  to  generalize  once 
more,  in  order  to  obtain  processes  simpler  and  more  direct. 
Indeed  Mathematics  might  be  called  a  '^  lazy  man's  de- 
light,'' because  it  is  continually  devising  methods  to  make 
our  work  shorter  and  less  laborious.  At  the  same  time, 
whenever  a  new  one  of  these  intellectual  "  tools  "  is  in- 
vented, it  is  found  to  be  wonderfully  adapted  to  the  solu- 


INCREMENT  AND  DERIVATIVE  23 

tion  of  new  problems.    Among  these  intellectual  tools  the 

simple  derivative,  -p,  is  perhaps  the  most  wonderful.* 

We  shall  now  proceed  to  simplify  the  process  of  finding 
a  derivative,  considering  first  algebraic  functions.  The 
operations  of  Algebra  are^ddition  (including  subtraction), 
multiplication,  division  (which  may  be  reduced  to  that  of 
multipUcation  by  the  use  of  negative  exponents),  raising 
to  powers,  and  extraction  of  roots  (reducible  to  raising  to 
powers  by  the  use  of  fractional  exponents).  We  need  only 
consider  then  differentiation  as  applied  to  functions  involv- 
ing addition,  multiplication  and  raising  to  powers. 

7.  The  Derivative  of  a  Sum.  Let  y  =  u  +  v  +  w+... 
where  Uj  Vj  w  .  .  .  are  functions  f  of  an  independent  va- 
riable X.  Give  X  an  increment  A  x,  thus  u,  Vy  w  .  .  .  will 
receive  increments  A  i^,  A  ?;,  A  k;  .  .  . ,  and  y  an  increment 
A  2/.    Then 

2/  +  A2/  =  '?^  +  A'W  +  z;  +  Ay  +  if;  +  Ai/;+-  •  • 

A^  =  (i/  +  Ai^  +  2;  +  Ay  +  ^-|"Aii;+  •  •     ) 
—  {u -\-  V  -{-  w  -{-  .  .  .) 

or         A2/  =  A2i  +  Az;  +  Atf;+... 

A  y  __  A  u   .    A  V   I    A  w;   , 
Ax       Ax       Ax       Ax 

and   limit      Al  =  limit     ( AJ^  +  ^  +  AE  +        \ 
Ax  =  o   Ax        Ax  =  o\Ax        Ax        Ax  / 

=  limit    A:?f  + limit      Al  + 

Ax  =  0     Ax  Aa^=0Ax 

limit     A^  + 

Ax  =  o   A  X 

*  The  weight  of  the  evidence  seems  to  give  to  Sir  Isaac  Newton  (1G42-1727  A.D.) 
the  honor  of  the  invention  of  the  calculus. 

t  Continuous  functions.    The  word  continuous  will  be  dropped  henooforth. 


24  CALCULUS  AND  GRAPHS 

or,  as  we  now  write, 

dy  _  d(u  +  V  +  w  +  . .  . )  _  du   .    d^   .    dw  .  .  . 

dx  dx  dx       dx      dx 

which  is  memorized  briefly  as  ^^  The  derivative  of  a  sum 
is  equal  to  the  sum  of  the  derivatives/'  * 

8.  The  Derivative  of  a  Product.    Let  y  =  uv 

where  u  and  v  are  functions  of  x.     Proceeding  as  above 

y  +  Ay  =  (u+  An)  (v  +  Av) 

A  y  =  (u  -\-.A  u)  {v  -}-  A  v)  —  uv 

=  vAu  +  uAv  +  AuAv 

Ay         Au  ,       Av   ,    .      Av 
Ax    Ax  Ax  Ax 

limit     4^  =  limit     L^  +  uAl+^u4A 
Ax  =  o   Ax        Ax  =  o\    Ax  Ax  Ax) 

=  Umit      y  Aif  +  Umit      u-^  + 

Aa;=0  Ax  Aa:  =  0  Ax 

limit       Au^ 
Ax  =  o  Ax 

or 

dy       diuv)         du   ,       dv  (3) 

dx         dx  dx  dx 

memorized  briefly  as,  '^  the  derivative  of  the  product  of 
two  factors  equals  the  second  factor  times  the  derivative 
of  the  first,  plus  the  first  factor  times  the  derivative  of  the 
second. '* 

*  This  is  not  self-evident,  nor  does  experience  lead  us  to  think  it  true.  The  square 
of  a  sum  is  not  the  sum  of  the  squares;  the  square  root  of  a  sum  is  not  the  sum  of  the 
square  roots;  the  sine  of  a  sum  is  not  the  sum  of  the  sines,  etc. 


INCREMENT  AND  DERIVATIVE  25 

Note  in  (3)  that  lii^^i^     A  u—--  =  0  because    A  u  ap- 
proaches  zero  as  A  x  does. 

9.  The  Derivative  of  a  Power.    Let  y  =  vJ" 
where  w  is  a  function  of  x.    As  above 

y-\-^y  =  {u+^  u)"' 

Ay  =  {u+  Aur  —  vT 

=  u^+nu-'  Au+  ^i^^)  u--'Alf+  .  . . 
+Au  —  w" 


Ax  Ax  1.2  Ax 

+  Az^       -r- 

A  X 
limit     Al=  limit      n?^^-'— + 

Ax  =  0   A  X  Aa:  =  0  A  X 

limit    "J!^:^)  ^«-2  A„4^  +  . . . 

Ax  =  0        1.2  Ax 

or  dy  _  d}£  _       ^-i  ^  /j^x 

dx        dx  dx 


In  particular  if  ti  =  x 


§-"  =  nz"-'  (5) 

dx 


Theorems  (4)  and  (5)  have  been  proved  for  positive  in- 
tegral values  only  of  n.  They  are  true  for  fractional  and 
negative  exponents,  but  the  proof  is  omitted. 

We  now  have  the  following  eight  theorems,  or  formulas, 
of  differentiation. 


dx 

dx      dx      dx 

diuv)         du  ,       dv 

-3-^  =  ^3~  +  '^  J" 
dx           dx          dx 

du^           n-i  du 
dx                 dx 

dx^               n-l 

-7-  =  nx^ 
dx 

dx  _  . 
da;  " 

(6) 


26  CALCULUS  AND  GRAPHS 

10.  Collected  Formulas  of  Dififerentiation. 

dw 
1. 


u. 


111. 


IV. 


dcu         du  ^     j_ 

VI.    -7—  =  c  -r-;  c  a  constant. 
dx  dx 

vii.    -7-  =  0;  c  a  constant, 
da; 

tU         du         dv 

d-       v~z u-— 

V  dx  dx 

Vlll.       -:-     =    — 


The  proofs  of  the  last  four  being  left  to  the  student.    To 
these  we  add, 

d  sin  u                du  iii.     d  sec  i^                ^          du 

1.       ; =   cos  U  -—•  ; =   SCC  U  tSLli  U  -y- 

dx                   dx  dx                             dx 

d  tan  u          ^     du  .        d  cos  u              .       du 

11.       ; =  SeC"^  U  -r-  IV.      — =  —  Slli  u  -r 

dx                     dx  dx                         dx 

d  cot  u  c,    du                             .^v 

V.     — =  —  csc^u  -7-                              (7) 

dx  dx 

d  CSC  u  ,      du 

VI.     — - —  =  —  cscu  cot  u  -J- 

ax  ax 


INCREMENT  AND  DERIVATIVE 


27 


To  prove  the  formula  — - —  =  cos  m  —  we  proceed  as 

ClX  CLX 

follows:  Draw  a  circle  (Fig.  11)  with  centre  at  0  and  with 
Y 


FlQ.    11 

radius  unity  (so  taken  for  convenience).  Let  the  angle 
LO  K  =  Xj  measured  in  radians,  and  angle  LOB  =  x  + 
Ax.  Draw  KN  and  BM  perpendicular  to  OX  and  KP 
perpendicular  to  BM;  also  KT  tangent  to  the  circle  at  K, 
Obviously  the  angle  N KT  =  x,  NK  =  sin  x  and  MB  = 
sin  (x  +  Ax),  Also  arc  KB  =  A  x  since  the  angles  are 
expressed  in  circular  measure,  and,  by  trigonometry,  arc 
KB  =  A  X  times  radius  OK  =  Ax,    We  thus  have 

PB  =  A  sin  a: 
PB         A  sin  X 


arc  KB 


A  X 


limit         P^     =  limit       ^^^^^  =  ^^i^^ 
KB'^arcKB         Ax  =  o      Ax  dx 

Also 

PB 


(a) 


chord  KB 


COS  PBK 


28         •  CALCULUS  AND  GRAPHS 

PR 
limit    _i_e —  =  limit  cos  PBK 
KB  =  o  chd.  KB 

=  coslimit  PBK 

=  cos  N KT  =  cos  a;  (b) 


But 


limit =  limit 


arc  KB  chd.  KB 


since,  when  the  point  B  approaches  the  point  K  as  its 
limiting  position,  chord  KB  and  arc  KB  become  more 
and  more  nearly  equal.    As  it  is  usually  expressed 

,.    .,  chord  KB       ^  ^ 

limit =  1  * 

arc  KB 

Therefore,  from  (a)  and  (b)  we  have 

d  sin  X 

— ; —  =  cos  X 
ax 

We  see  again  that,  u  being  a  continuous  function  of  x, 

A  sin  16  _  A  ^inu     A  u 
Ax  Au        Ax 

and 

limit      ^^^^^  =  limit       ^  "^"  ^  .  limit     A^ 

Aa;  =  0         Ax  At*  =  0         Au  ^x  =  0    Ax 

or  d  sin  u  du 

— - —  =  cos  u  — 

dx  ax 

To  prove  the  formula 

d  cos  u  .       du 

— ; =  —  sin  1/  -— 

dx  dx 

*  This  statement  is  not  proved  but  is  taken  as  sufficiently  obvious  for  our  purpose. 


INCREMENT  AND  DERIVATIVE  29 

we  might  proceed  as  in  the  case  of  the  derivative  of  the 
sine,  using  Fig.  11,  or  as  follows: 


d  cosu 


dsin(  —  —  u)  (tt         \d(-z  —  u] 


dx  dx  ^  dx 

d( —  u)  .       du 

=  sin  u  -^ =  —  sin  u  -7- 

dx  dx 

Also,  since  tan  u  = ,  sec  u  = etc.,  the  remain- 
cos  u                 cos  u 

ing  formulas  of  group  (7)   can   be  proved  by  using  the 

derivatives  of  sine  and  cosine  together  with  the  formulas 

of  group  (6),  Art.  10.    These  proofs  are  left  as  an  exercise 

for  the  student. 

By  means  of  a  few  examples  we  will  illustrate  the  use  of 

the  formulas  of  differentiation. 

Example  1.        y  =  2x^+Sx  —  2  x^^^ 

dy  ^  d(2  x^  +  3  X  —  2  x^h 
dx  dx 

which  is  the  derivative  of  a  sum.    Therefore,  by  (6) — i 

di  ^  d2x^       dSx  _ d2_x^ 
dx  dx  dx  dx 

and  by  (6) — vi 

dy  __     dx^   .      dx       ^dx^^^ 

di-^l^^^d~x~^lbr 

whence  by  (6) — iv  and  v 

^  =  2.3x2+ 3.1- 2.  H^"'/'  =  6x-'  +  3-47o 
dx  x^/2 

Example  2,  x~  +  2 

or  1/  =  (x-  +  2)  (x  —  1)-^ 


30 


CALCULUS  AND  GRAPHS 


|  =  <._„--*i±i'  +  («=+2,*^\y(e,-ii 


-(i-I)- 


S  +  Tfl+'-^  +  ^'l-*^-"" 


d(x  —  1) 


by  (6)-i 

=  (x-ir'{2x  +  Q)  +  (x'+2) 

by  (6) — iv;  vii  (6) — i 

=  (x-  1)"^  2x-(x''+2)  {x-  l)-2  (1  -  0)  by  (6)— v;  vii 


dx 

(6)- 

-iii 

[-(.- 

-ir 

_dx 

d 
d 

dy 


2x 


x'  +2        x'  —  2x 


dx      x  —  l       {x—iy  {x  —  l)^ 

Example  8.  y  =  2  sin^  3  x 

dy       ^d  sin^  3x       nn-o     c?sin3x 

-r-  =  2 =  2 . 2  sm  3  a; ; 

dx  dx  dx 


(6)-vi 


(6)-iii 


=  4  sin  3  a:  cos  3  x  — . —  =  4  sin  3  x  cos  3  x  .  3 
dx 

(7)-i  (6)-vi 

=  12  sin  3  X  cos  3  a;  =  6  sin  6  x 

Example  4.  y  —  x^  tan  2  x 

dy       .      ^     dx^   .      „  d  tan  2  a:  (6) — ii 

-r  =  tan  2  a;  -; — h  x^ 

dx  dx  dx 

=  tan  2  a;  .  2  a:  +  a;^  sec^  2  x 


d2x 
dx 


(6)-iv 


(7)-ii 


INCREMENT  AND  DERIVATIVE 

=  2  X  tan  2  a:  +  x-  sec^  2  x  .  2 
=  2x  (tan  2x-\-  X  sec^  2 x) 

Example  5,      x^y^  +  2xy  =  7  x'^ 

d(xY  +  2  xy)  ^  dlx'' 
dx  dx 

dx     "^      dx  dx 


31 

(6)-vi 


(6)-i 

y'  — — \-  X-  - — \-  Z 
dx  dx 

(6) — ii  and  vi 


(6)-vi 

dx   .       dy 
dx  dx 


'   =  7.2x 


y^  .2X  +  X''  .Zy^^+2y^-2x'^y 


dx 
(6)— iv;    iii;    v 


dx 


Ux 


2  xi/^  +  3  x2|/2  ^+2y  +  2x^=Ux 
ax  ax 

dv 
which  is  an  algebraic  equation  of  the  first  degree  in  -^' 

dx 

Whence 

(3  xY  +2x)^  =  Ux  —  2y  —  2xy^ 
dx 


dy  _  14  a:  —  2y  —  2  xy^ 

dx  ~        2x4-3  xY 


Note:  In  equations  such  as  that  in  Example  5  the  y  depends  upon  x 
for  its  value  and  is,  therefore,  a  function  of  x,  but  as  the  value  of  y  in 
terms  of  x  is  not  exactly  expressed  but  merely  implied,  we  say  that  y 
is  an  implicit  function  of  x.  When  y  is  exactly  expressed  in  terms  of 
X  we  call  y  an  explicit  function. 


32  CALCULUS  AND  GRAPHS 

EXAMPLES 

In  each  of  the  following  examples  find  the  derivative  of  the  function. 


1.  Sx^  —  2x-\-^ 

5. 

{x^ 

+  2)3 

9.  sin  X  cos  x 

2.  x^  —  x^  +  Sx 

6. 

(3- 

—  ^3)2 

10.  sin  X  tan  re 

3.  ix  +  3)3 

7. 

(x^ 

+  X)V3 

11.  cosx  +  sec  2  a; 

4.  (2  +  x)-2 

8. 

(X 

—  2x2) 

V^ 

12.  sin  2  X  —  tan  3  x 

n        1       -^      ^ 

; 

17.  3  x2.7 

lo.                -+-  ■ 

u+2       M— 2 

14.?           ^ 

S      S  +  7 

18.  5X-6-02 

^^'  y(y^  +  2) 

19.     1^x3-14 

16.       ' 

20.  —  2a;-2V« 

2/  — r 

21.  ^2  +  2/2  =  16  22.  4  x2  +  5  2/2  =  20 

23.  4  x2  —  5  ?y2  =  20  24.  xy  =  100 

25.  x?/  +  X  —  2  2/  +  7  =  0  26.  X2/  —  3  X  —  2/  =  4 

27.  sin2  (2  x  +  I)  28.  cot3  (x  —  |) 

29.  cos  x  .  sec  (2  X  +  ^)  30.  sin  x  cos  2  x  tan  3  x 


CHAPTER  III 

APPLIED  MEANINGS  AND  USES  OF  THE  DERIVA- 
TIVE. RATE  OF  CHANGE.  VELOCITY.  ACCEL- 
ERATION 

11.  Rate  of  Change.    Velocity.    Acceleration.    Suppose 

there  is  a  quantity,  Uj  which  changes  in  value  as  time 

elapses,  and  suppose  that  a  change  (increment)  of  time 

A  u 
A  t  causes  an  increment,  A  u,  in  u.    Then  -r— -  will  be  the 

A  ^ 


average  rate  of  change  of  u  during  the"Time  A  t.  For  ex- 
ainple,  suppose  theTempefature  of  theToonTtiasrisen  from 
40°  to  70°  between  9:00  A.  M.  and  11:00  A.  M.  Then  the 
temperature  has  risen  at  the  average  rate  of  15°  per  hour 
or  .25°  per  minute. 

Au       70  —  40       30       ,^ 

=    -r-     =    15. 


A^         11  —  9         2 

Can  we  say  that  at  10:00  o'clock  the  temperature  was 
changing  at  that  same  rate?  Obviously  not,  because  at 
9:55  o'clock  the  temperature  may  have  risen  to  80°,  the 
heating  steam  may  have  then  been  turned  off  and  the 
temperature,  stationary,  perhaps,  for  5  minutes,  have  fallen 
between  10:00  and  11:00  A.  M.  from  80°  to  70°.  Thus 
the  temperature  at  10:00  A.  M.  may  have  been  actually 
falHng,  instead  of  rising  at  the  rate  of  .25°  per  minute. 
Again  suppose  we  walk  to  school  from  home,  a  distance  of 
3  miles,  in  45  minutes.  Our  average  rate  of  walking,  our 
average  speed,  the  average  rate  of  change  of  the  distance, 
would  be  3  miles  -r-  |  hours  =  4  miles  per  hour.  Can  we 
say  we  crossed  Avenue  A  at  10th  Street  at  4  miles  per 

33 


34  CALCULUS  AND  GRAPHS 

hour?  Probably  not.  If  the  automobiles  were  numerous 
our  speed  (rate  of  change  of  distance)  was  more  likely  to 
be  six  or  eight  miles  per  hour.  How  then  can  we  find  the 
rate  of  change  of  a  quantity  at  a  given  instant,  the  in- 
stantaneous rate  of  change,  and  what  is  meant  by  in- 
stantaneous rate  of  change?  Suppose  Avenue  A  is  52.8 
feet  broad  and  we  crossed  it  in  6  seconds.  Our  average 
speed  across  the  avenue  was  6  miles  per  hour.  But  when 
two-thirds  the  way  across  we  may  have  speeded  up  to 
dodge  a  car,  so  that  the  last  17.6  feet  were  covered  in  one 
second.  Our  average  speed  over  this  distance  was,  there- 
fore, 12  miles  per  hour.  How  can  we  find  the  speed  at 
whichjve  mounted  thp.  nnrV>  ot]  f^ji^  f^^  ^^^f^  ^^  ^he  avenue? 
We  see  how  it  is  done.  By  taking  the  distance  and  time 
shorter  and  shorter  we  find  the  average  speed  through  a 
shorter  and  shorter  distance  and  time,  and  by  taking  the 
distance  and  time  short  enough  we  obtain  the  average 
speed  for  a  time  and  distance  so  short  that  we  can  call  it 
the  instantaneous  speed.    Thus: 

Instantaneous   speed  =  limit  — -. as   distance   and 

time 

time  approach  zero. 

In  mathematical  symbols 

Speed  =  V  =  l™it       Aj  ^  ^^  rg^ 

and,  in  general,  if  u  is  any  quantity  which  changes  in  value, 
the  instantaneous 

Rate  of  change  of  i^  =  ^^^^^  ^   ^  =  p  (9) 

^  LX  L  (XL 

Obviously,    A  t  being  taken  positive,  if  -r—  is  j  ^ 


RATE  OF  CHANGE  35 


so  also  is  A  u,  and  therefore  u  is 


increasing 
decreasing 


But  if 


-7 —  is  always 
A  t 


positive 
negative 


then  also  must  "^^I't       — — 


du  I 

-—  be 
dt 


positive  1       ,  j.y.  j_  'I-  du  ,    [positive  ] 

hand  we  can  say  that  if  -rr  is  i  ^.      \ 

negative]  dt        [negative] 


^  ,  increasing 

u  must  be  an  i  , 

\  decreasing 


function. 


It  must  be  noted  that  speed  is  merely  a  special  case  of 
rate  of  change;  that  is,  the  rate  of  change  of  a  distance. 
It  occurs  so  frequently  that  a  special  name,  speed,  is  given 
to  it.  Another  rate  of  change,  of  frequent  occurrence,  to 
which  a  special  name  is  given  is  the  rate  of  change  of  the 
speed.  The  rate  of  change  of  speed  is  called  acceleration. 
In  symbols 

Acceleration  =  f  =  -j.  (10) 


which  may  also  be  written 


^=-ir-S  <"> 


d^ 
dx 
respect  to  Xj  and  means  the  derivative  of  the  derivative 


In  general  -j^^  is  called  the  second  derivative  of  u  with 


12.  Angular  Velocity  and  Acceleration.  Another  impor- 
tant instance  of  rate  of  change  is  that  which  arises  in  the 
case  of  a  rotating  or  revolving  body.  Suppose  a  wheel 
turns  on  its  axis,  making  100  revolutions  per  minute. 
(Fig.  12);  then  100  revolutions  per  minute  =  2007r  radians 


38 


CALCULUS  AND  GRAPHS 


per  minute  is  called  the  angular  velocity.  In  this  case  the 
angular  velocity  is  constant,  but  if  the  wheel  is  coming  to 
rest  because  of  the  friction  of  the  air  and  of  its  axle,  the 
angular  velocity  will  not  be  constant.  Reasoning  as  in 
previous  instances  we  may  say  that 


Fig.  12 


Ai 


=  average  angular  velocity  during  the  time  A  t 


and 


^  ^^^^^^  Q  -TT  =  angular  velocity  at  any  instant  =  co.  (12) 


Similarly 


_<1) 


-77  =     ^    „     =  -r;^  =  tt  =  the  rate  of   change  of   the 
at  at  dt^ 


angular  velocity;  called  angular  acceleration  (13) 

As  we  learned  in  Trigonometry,  the  angle  6  being  ex- 
pressed in  circular  measure  or  radians,  (see  Fig.  12). 


RATE  OF  CHANGE  37 


'  or  s  =  au 
a 


whence 


ds         dd.  ,^.. 


ds 
where  —  is  the  speed  of  a  point  on  the  rim  of  a  wheel  of 
dt 

radius  a.     The  motion  of  the  point  is  in  the  direction  of 

the  tangent  to  the  curve  at  any  instant. 

13.  Components  of  Velocity.  At  the  point  B  (Fig.  12) 
let  BP  represent  in  length  and  direction  the  velocity  of  the 
point  at  any  instant.  Then,  as  we  have  learned  in  our 
study  of  Physics,  BP  can  be  resolved  into  two  components, 
CP  parallel  to  OX  and  BC  parallel  to  OY. 

From  the  figure 

CP  =  BP  sin  e     and     BC  =  BP  cos  6 

where  CP  and  BC  are  the  rates  of  change  of  distances 
X  and  y  parallel  to  OX  and  OY  respectively.  We  can, 
therefore,  write 

dx  ds   .    ^         ,  dij       ds        ^ 

or 

Vx  =  —  v  sin  d  Vy  =  V  cos  0  (15) 

It  is  obvious  that 

vl  +  vl  =  v^  (sin2  d  +  cos^  6)  =  v\ 

That  is,  the  speed  of  a  particle  in  its  path  equals  the  square 
root  of  the  sum  of  the  squares  of  its  component  speeds 
parallel  to  the  axes  of  x  and  y. 


38 


CALCULUS  AND  GRAPHS 


Similarly  (Fig.  13)  if  a  particle  is  moving  along  a  path 
in  a  system  of  rectangular  coordinates,  the  distance  from 
some  initial  point,  A,  being  represented  by  S  and  the 


Fig.  13 


tangent  to  the  path  at  P  making  an  angle  ^  with  OX 

we  can  represent  by  PT  the  velocity  of  the  particle  at  P, 

ds 

-r-,  and  by  PR  and  RT  the  velocity  components  parallel 

to  OX  and  0  Y  respectively.    We  then  have 

PR  =  PT  cos  (I)    and    RT  =  PT  sin  cj) 


or 


dx       ds 


,  dy       ds    .     , 
and  "i"  =  —  sm  d) 
dt       dt 


(16) 


or 


Vx  =  V  COS  ^ 


and  Vy  —  V  sin  ^ 


which  are  the  same  relations  as  (15)  since  (Fig.  12),  <[)  = 
90°  +  d,    cos  ^  =  —  sin  0,    sin  ^  =  cos  0. 

For   convenience  of  reference  the  various  formulas  of 
Arts.  11  and  12  are  here  assembled. 


RATE  OF  CHANGE 


39 


—  =  rate  of  change  of  any  variable  u  (with  respect  to  time,  t) 
at 


ds 


dt 
dh 
de 
dd 

dt 
d^ 

df" 
dx 


=  V  =  velocity  or  speed. 


=  —  =  /  =  acceleration 
dt 

=  0)  =  angular  velocity. 

dco 


(17) 


dt 


=  a 


angular  acceleration. 


ds 


,        —  cos  (i>  or  Vx  =  V  cos  ^ 
dt       dt 

dy       ds   .     ,  .     . 

-^  =  -r  sm  ^  or  Vy  =  V  sin  ^ 
dt       at 


horizontal  and 
vertical  components 
of  velocity. 


(d_sY  ^  (dxV       (dyV 
\dtj        \dtj  "^  \dtj 


or  v^  =  vl  -\-  v^. 


Example  1.  A  body  moves  along  a  straight  line  running 
East  and  West  so  that  its  distance  from  a  fixed  point  0  is 
given  by  the  equation  s  =  t^  —  3  ^^  +  2,  s  being  distance 
and  ty  time.  What  is  its  speed  at  any  instant?  When 
<  =  3?  When  will  the  body  be  stationary?  When,  how 
long  and  how  far  will  it  move  east?    West? 


ds 
v=-j-  =  St^  —  6^  and  when  r 
dt 


0, 2;  =  0;  when  t  =  3yV  =  9. 


When  the  body  is  stationary  the  speed  is  zero,  therefore 

3  ^2  _  g  ^  _  0,  ^  =  0,  2 
Tabulating  these  results 


t 

0 

1 

2 

3 

ds 
dt 

0 

— 

0 

+ 

40  CALCULUS  AND  GRAPHS 

Thus  for  the  first  two  seconds  ~  is  negative  and  since  A  t 

is  positive  (time  increases)  s  is  decreasing  and  the  body 

ds  . 
moves  west.    After  two  seconds  —  is  positive,  s  increases 

and  the  body  moves  east.  The  body  starts  at  s  =  2  (put 
t  =  Oin  s  =  f  —  Sf-+2).  When  t  =  1,  s  =  0;  therefore 
the  body  moves  west  a  distance  of  two  units. 

Example  2.  A  spherical  balloon  is  being  filled  with  gas 
at  the  rate  of  2  cubic  feet  per  minute.  How  fast  is  the 
radius  of  the  sphere  increasing  when  the  radius  is  10  feet? 

V  =  ^  IT  r^ 

dv       .         n    ^dr        .        ^dr 
-=^7r.3r--=4rr-- 

-"  =  2  cu.  ft.  min.:    r  =  .10 
dt 

.  .  2  =  400  TT  -  ;    -  =  ^^^      ft.  mm. 
dt     dt       200  TT 

Example  3.  The  pressure  times  the  volume  of  a  gas  is 
equal  to  16.  The  volume  is  decreasing  at  the  rate  of  3  cu. 
cm.  per  second.  At  what  rate  is  the  pressure  changing? 
When  the  Volume  is  10  cu.  cm.? 

pv  =  16 


and 


dp   ,       dv 

'di  +  ^-dt'- 

=  0 

dv  _ 

dt-    ^ 

dv 
■  -'It- 

-3p  =  0 

dp  _Sp 

dt         V 

RATE  OF  CHANGE  41 

When  -^  16       ^  ^  , 

t;  =  10,     p  =  —  =  1.6     and 

V 

#  =  !:§=   4« 
dt       10       * 

Example  4-  A  metal  cube,  remaining  cubical,  expands 
under  the  influence  of  heat.  How  much  faster  does  the 
volume  increase  than  the  edge? 

..      dv       ^    odx  dt         o     o 

V  =  x'\     -T  =  3  .T-  —     or    -r-  =  3  X-, 
dt  dt  dx 

dt 

so  that  the  volume  is  increasing  3  x'  times  as  fast  as  the 

edge;  in  particular  when  a;  =  2,   12  times  as  fast;  a;  =  3, 

27  times  as  fast. 

The  interval  of  time  being  the  same  in  both  derivatives 

we  may  write  * 

dv 

-r-  =  —  =  Sx'j  the  rate  of  change  of  the  volume  with  respect 
dx      dx 

It 

dv 
to  the  edge  of  the  cube.    In  general  ^  may  mean  the  rate 

ax 

of  change  of  y  with  respect  to,  or  as  compared  with  x.  (18) 


Example  5.  A  wheel  of  radius  2  feet  makes  100  revolu- 
tions per  minute.    How  fast  is  a  point  on  the  rim  moving? 

A  V  A .'/  Ay 

*  T«     „«r,„.oi     ^  ^        ^  ^        limit        ^  ^        limit       A  ^        *"^^    A~^ 
♦In     general    ^^  =  ^.     ^.  =  o^,    =A-0A^=--^ 

dy  dv 

OT  -r-   =  J-.     Thus  3-  =  3".    Similarly  -f   =  t-  •  t"- 
dx        dx  dx       dx  dx       dz     dx 

dz  di 


42  CALCULUS  AND  GRAPHS 

What  are  the  x  and  y  components  of  velocity  when  the 

3  TT 

wheel  has  moved  through  — —  radians? 

rl  R 

Here    —r  =  100  rev.  min.  =  200  w  radians  minute. 
at 

s  =  ad  =  26       /.  ^f  =  24r  =  400  tt  feet  minute 
at  at 

3   TT 

and    is    the    same    for    all    angles.       Also   at    0  = 


4 


^^^        .     3  TT  400  w 

v^  =  —  400  TT  sm  — —  = y=- 

4  V2 

,  ^nn  ^  ^         400  TT 

and    2;jy  =  400  w  cos——  = 


Uy 


4  V2 


Example  6.  The  power  which  turns  the  wheel  of  Ex- 
ample 5  is  cut  off,  after  which  friction  stops  the  wheel 
according  to  the  law  6  =  200  it  t  —  f.  Find  the  angular 
velocity  and  acceleration,  the  velocity  and  components  of 
velocity  of  a  point  on  the  rim;  at  any  instant  and  after 
2  hours  from  the  time  the  power  is  shut  off.  In  what  time 
will  the  wheel  be  brought  to  rest? 

CO  =  -r-  =  200  TT  —  2t  rad.  min. 
dt 

a  =  -r-^  =  —  2  rad.  per  mm.  per  mm. 

v  =  ^  =  2^  =  4007r--4ift.  min. 
dt  dt 

v^  =  — «;  sin  0  =  —  (400  w  —  4  i)  sin  0  ft.  min. 

Vy  =  V  cos  6  =  (400  TT  —  4:t)  cos  0  ft.  min. 

When  ^  =  2  hr.  these  become  (note,  t  =  2  hrs.  =  120  min. 
gives  6  =  24000  tt  —  14400  =  60960  rad.  =  3,493,008°  = 
9702  X  2  7r+288°). 


RATE  OF  CHANGE  43 

CO  =  ^  =  200  TT  —  240  =  388  rad.  min. 
dl 

V  =  '^  =  400  TT  —  480  =  776  ft.  min. 
dl 

v^  =  —  776  sin  288°  =  —  776  X  (—  .951)  =  738  ft.  min. 

Vy  =       776  cos  288°  =       776  X  (      .309)  =  240  ft.  min. 

The  wheel  will  come  to  rest  when 

^  =  200  TT  —  2  ^  =  0  or  ^  =  100  TT  min.  =  5  hr.  14  min. 
dt 

EXAMPLES 

In  the  following  examples  in  which  s  is  distance  in  feet,  t  is  time  in 
seconds.  Find  (1)  velocity,  (2)  acceleration,  (3)  starting  point,  (4) 
stationary  points,  (5)  when  and  how  long  the  particle  moves  N.,  S., 
S.  W.,  etc.,  as  the  case  may  be,  (6)  how  far  it  moves  N.,  S.,  etc. 

1.  s  =2  +  3  ^2  —  /3      Motion  on  a  straight  line  running  east  and 

west. 

2.  s  =  t^  —  St^+2  Straight  line  N.  and  S. 

3.  s  =  7  —  5^  +  9^2  Straight  line  N.  E.  and  S.  W. 

4.  s  =  5  +  4  <  —  <2  Straight  line  N.  W.  and  S.  E. 

5.  s  =  t^  —  St^  '  Straight  line  E.  and  W. 

6.  s  =  3  «2  __  ^3  Straight  line  N.  and  S. 

If  a  body  falls  near  the  surface  of  the  earth  its  motion  is  represented 
by  s  =  16  ^2  if  [I  falls  from  a  state  of  rest;  by  s  =  16  ^^  —  ^^^  jf  [^  j^^^j 
an  initial  upward  velocity  vo,  where  s  is  in  feet  and  t  in  seconds. 

7.  Find  the  velocity  and  acceleration  of  a  body  falling  from  rest, 
at  the  end  of  t  seconds;  2  seconds,  5  seconds. 

8.  Find  the  velocity  and  acceleration  of  a  falling  body,  with  an 
initial  velocity  downward  of  3  feet  per  second,  at  the  end  of  i  seconds; 
3  seconds,  10  seconds. 

9.  A  body  at  rest  1600  feet  above  the  earth  falls  to  the  earth.  With 
what  velocity  will  it  reach  the  earth's  surface? 


44  CALCULUS  AND  GRAPHS 

10.  If  the  body  of  question  9  is  thrown  downward  with  a  velocity  of 
20  feet  per  second,  with  what  velocity  will  it  reach  the  earth? 

11.  A  body  is  thrown  vertically  upward  with  a  velocity  of  320  feet 
per  second.    How  long  and  how  high  will  it  rise? 

12.  From  what  height  must  a  body  be  dropped  that  it  may  strike 
the  ground  with  a  velocity  of  200  ft.  per  second?  In  what  time  will  it 
reach  the  ground? 

13.  With  what  velocity  must  a  ball  be  thrown  vertically  upward  to 
rise  to  a  height  of  320  feet? 

14.  What  is  the  speed  of  a  body  falling  from  rest,  at  the  end  of  2 
seconds?  If  this  speed  remained  constant  for  one-tenth  second  how 
far  would  the  body  fall?  How  far  does  the  body  actually  fall  in  the 
one-tenth  second?  What  is  the  error,  and  the  percentage  error,  in 
assuming  the  speed  constant?  Answer  the  same  questions  for  one- 
hundredth  of  a  second. 

15.  The  ends  of  a  trough  6  feet  long  are  equal  isosceles  right  triangles 
having  the  hypothenuse  horizontal.  A  horse  is  drinking  the  water  at 
the  rate  of  a  cubic  foot  per  minute.  How  fast  is  the  level  of  the  water 
sinking  when  the  water  is  18  inches  deep? 

16.  A  trough  is  in  the  shape  of  a  right  prism,  with  its  ends  equilateral 
triangles.  The  length  of  the  trough  is  10  feet.  It  contains  water,  which 
leaks  at  the  rate  of  1  cubic  foot  per  minute.  Find  in  inches  per  second 
the  rate  at  which  the  level  is  sinking  when  the  depth  of  the  water  is 
3  inches. 

17.  A  vessel  containing  water  is  in  the  form  of  an  inverted  hollow 
cone  with  vertical  angle  90°.  If  water  flows  in  at  the  rate  of  1  cubic 
foot  per  minute,  at  what  rate  is  the  level  of  the  water  rising  when  the 
depth  is  2  feet? 

18.  In  an  hour  glass,  with  vertical  angle  90°,  the  sand  is  flowing 
from  the  upper  cone  to  the  lower  at  the  rate  of  x  cubic  centimeters  per 
second,  where  x  is  the  height  of  the  sand  in  the  upper  cone.  How  fast 
is  the  level  of  the  sand  sinking  when  the  height  of  the  sand  is  1  centi- 
meter?   Does  the  level  sink  faster  or  slower  as  the  sands  run  out? 

19.  A  current  C  of  electricity  is  changing  according  to  the  law  C  = 
20  +  21  ^ —  l^t^,  where  t  is  in  seconds.    The  voltage  V  is  such  that 

V  =  RC  +  L^,  where  R  =  0.5  and  L  =  0.01.    Find  V  when  t  =  2. 
at 

How  fast  is  the  voltage  changing  at  any  time?    When  ^  =  2?    Is  the 
voltage  increasing  or  decreasing  when  t  =  .73  sec?    Why? 


RATE  OF  CHANGE  45 

20.  The  coefficient  of  cubical  expansion  of  a  substance  at  temperature 
d°  is  the  rate  of  increase  of  volume  per  unit  increase  of  temperature. 
The  volume  (c.c)  of  a  gram  of  water  at  0°  C  is  given  by  F  =  1  + 

dV 
aid  —  4)2,  when  a  =  8.38  X  lO"*.     Find  —r  and  hence  get  the  co- 
rf fj 

efficient  of  cubical  expansion  of  water  at  0°  C.  and  at  20°  C. 

21.  The  specific  heat  of  a  substance  at  temperature  0°  is  the  rate 
of  increase  of  Q  per  unit  increase  in  0,  where  Q  is  the  number  of  heat 
units  required  to  raise  the  temperature  of  1  gm.  from  some  standard 
temperature  to  0.  The  total  heat  required  to  raise  the  temperature 
of  1  gm.  of  water  from  0°  to  0°  C.  is  given  by  Q  =  0  -\- 2  X  lO'^  0^  + 
3  X  10-7  0\    Find  the  specific  heat  of  water  at  80°  C. 

22.  For  a  diamond  the  formula  is  (see  Ex.  21) 

Q  =  0.0947  0  +  0.000497  0^  —  0.00000012  0\ 
Find  the  specific  heat  of  diamond  at  80°  C. 

23.  A  wheel  iii  t  seconds  rotates  through  5  ^  -|-  4  ^^  radians  from  some 
standard  position.  Find  its  angular  velocity  and  acceleration  after 
5  seconds. 

24.  A  fly  wheel  is  making  100  revolutions  per  minute.  When  the 
power  is  cut  off  friction  brings  the  wheel  to  rest  according  to  the  law 
0  =  200  izt  —  20  X  ^2  Find  the  angular  velocity  and  acceleration 
at  any  minute  after  the  power  is  cut  off;  when  t  =  2  min.  When  will 
the  wheel  come  to  rest? 

25.  A  wheel  of  radius  10  inches  makes  100  revolutions  per  minute. 
Find  the  velocity  components  of  a  point  in  the  rim  one  second  after 
it  crossed  OX. 

26.  A  particle  moves  in  a  curve  given  by  the  equations  a;  =  5  cos  ^ 
2/  =  5  sin  tj  where  t  is  the  time.  Prove  that  the  path  is  a  circle.  (Give 
values  to  t,  find  x  and  y,  and  plot  the  points.)  Find  the  x  and  y  com- 
ponents of  velocity  when  <  =  1,  <  =  2  t. 

27.  The  path  of  a  moving  particle  is  given  hy  s  =  t^  —  3  ^2  [^  is 
known  that  when  t  =  3  the  particle  is  moving  in  a  line  making  an 
angle  of  45°  with  OX.  Find  the  velocity  and  the  velocity  components 
when  t  =  3. 

28.  Which  will  come  to  rest  sooner,  a  wheel  revolving  according  to 
the  law  0  =  I  —  t-  or  one  whose  law  is  0  =  t  -]-  2  cos  t?  How  much 
sooner? 


CHAPTER  IV 


APPLIED  MEANINGS  AND  USES  OF  THE  DERIVA- 
TIVE, CONTINUED.  DIRECTION.  MAXIMA  AND 
MINIMA.  DIFFERENTIALS,  APPROXIMATIONS, 
ERRORS. 

14.  Direction  of  a  Line.  Slope.  Let  us  consider  a  simple 
function  y  =  2x  —  4,  of  which  the  graph  is  shown  in  Fig. 
14  to  be  a  straight  Hne.  Suppose  we  wish  to  know  the  di- 
rection of  this  straight  Une.     To  determine  the  direction 

we  must  have  some 
Q/  reference    Kne,    just 

•^  /  as  in   getting  direc- 

tions on  the  surface 
of  the  earth  we  use 
for  reference  the 
earth's  axis  and 
equatorial  plane,  and 
X  thus  get  directions 
northwest,  southeast, 
etc.  Let  us  choose 
the  axis  of  x  as  our 
reference  line  and 
measure  the  direction 
of  our  line,  y  =  2  x 
—  4,  by  means  of 
the  angle  cj)  made  by 
the  line  with  OX, 
to  the  right  of  the  line  and  above  the  axis.  It  is  found 
convenient,  in  most  cases,  however,  to  use,  instead  of  ^  the 
tangent  of  ([>.    This  is  called  the  slope  of  the  Une.     Thus 

Slope  BAP  =  tan  cj)  (19) 

46 


Fig.  14 


THE  DERIVATIVE 


47 


Let  P  {x,  y)  be  any  point  and  Q  (x  +  A  a;,  y  +  A  y)  any 
other  point  on  the  line.  Draw  the  coordinates  of  P  and  Q 
and  draw  PR  parallel  to  OX,  The  angle  RPQ  is  equal  to 
([).    Therefore 


Slope  BAP  =  tan  (]>  =  tan  RPQ  =  ^  =  ^ — 

PRON  —  OM 


{ij  +  iiy)—y 
(x  -{-  A  x)  —  X 


Ay 
A  X 


(20) 


so  that  the  slope  of  a  straight  line  may  also  be  defined  as 
the  increment  of  the  ordinate  (  A  y)  divided  by  the  incre- 
ment of  the  abscissa  (Ax), 

Consider  next  the  function  y  =  x"  of  which  the  graph 
(Fig.  15)  is  a  curved  Une,  whose  direction  changes  at  every 


point.  What  is  meant  by  the  direction  of  this  Hne  at  the 
point  P?  As  the  point  Q  approaches  nearer  and  nearer  to 
P  the  secant  PQ  approaches  nearer  and  nearer  in  position 


48  CALCULUS  AND  GRAPHS 

to  the  tangent  line  PT,  and  the  angle  XAP  between  the 
axis  of  X  and  the  secant  approaches  nearer  and  nearer  to 
the  angle  ^  between  the  axis  of  x  and  the  tangent  line. 
By  taking  the  point  Q  as  near  as  we  please  to  P  we  can 
make  the  position  of  the  secant  PQ,  and  the  angle  of  the 
secant,  approach  as  near  as  we  please  to  the  position  and 
angle  of  the  tangent  line  respectively.*  Thus,  to  determine 
the  direction  of  a  curve  we  have 

Slope  of  curve  at  P  =  slope  of  tangent  line  to  curve  at  P  = 
limit  of  slope  of  secant  PQ  as  Q  approaches  P  as  a  limit; 
or,  in  symbols. 

Slope  of  curve  at  any  point  =  l™i^       ^  ^  ^  (21) 

For  the  particular  function  sketched  in  Fig.  15 

Slope  =  -7-  =  2  a:. 
dx 

At  the  point  x  =  1,  y  =  \,  slope  =  2;  at  the  point  (5,  25) 
slope  =  10,  etc. 

By  virtue  of  (18)  p.  41  the  slope  of  a  curve  may  also  be 
considered  as  the  rate  of  change  of  the  ordinate  with  re- 
spect to  the  abscissa. 

Example  1,  Find  the  slope  of  the  tangent  to  the  curve 
y  =  x^  —  a;^  at  any  point  and  at  the  point  (2,  12);  also  the 
slope  of  the  chord  joining  (2,  12)  to  (2.1,  15.04). 

^  =  4:x^  —  2x;sit  (2,  12),  slope  =  4(2)'^  —  2(2)  =  28 

Q/X 

A2/_  15.04— 12       3.04 

Ti  ~    2.1-2    -^  -  ^^-^ 

Example  2.  Find  the  slope  of  the  curve  s  =  ^^  —  3  at  the 
point  ^  =  2,  s  =  13. 

*  The  student  should  show  this  by  a  carefully  drawn  figure. 


THE   DERIVATIVE  49 

^  =  4  t')  at  (2,  13),  slope  =  4(2)^=  32 
at 

It  should  be  noted  that  if  s  represents  distance  and  t  time, 
the  slope  of  the  curve  in  this  case  is  the  speed  of  the  moving 
body  in  its  path, 

EXAMPLES 

Find  the  slope  of  each  of  the  following  curves  at  any  point,  and  at 
the  points  named,  and  also  the  slope  of  the  chord  joining  the  named 
l)oint  to  the  point  whose  abscissa  is  greater  by  one-tenth. 

1.  ?/  =  2a;2  — 3  (2,5) 

2.  2/  =  4  — 3.c2  (—1,3) 

3.  2/  =  a;3  +  2  .r  (2,  12) 

4.  2/  =  x3  ~  a;2  (3,  18) 

Has  the  particle  moving  in  the  path  given  below  (s  =  distance, 
/  =  time)  a  greater  speed  when  ^  =  1  or  when  t  =  2?  Verify  your 
result  by  sketching  the  graph. 

9.  s=3^2_4^  10.  s  =  i2  — 1  11.  si  =  12 

12.  s  =  1  —  t^  13.  s  =  ^3  _  4  14.  s  =  i  —  «2 

Find  the  slope  of  each  of  the  following  curves  at  the  points  named. 

18.  y  =  tan  2",        x  =  0,  ^• 

19.  y  =  sec  S  Xj      x  =  0,  ^• 

20.  2/  =  CSC  ^,        X  =  0,  |- 


5.  xij  =  10 

(2,5) 

6.  x^y  =  20 

(2,5) 

7.  s  =  t'~3P 

(3,0) 

8.  s  =  3<2_t3 

(3,0) 

15.  y  =  sin  2  x, 

X 

•H.        Tt       Tt 

=   2»    Tj   G- 

16.  ?/  =  cos  2  X, 

X 

-    2f  T>   G- 

17.  y  =  tan  x, 

X 

=  0,1. 

15.  Maxima 

and 

Minima 

x^    ,    Sx 

2 

1  /^\                T 

Consider    the    function 
y  =  —^  +  ^^  (Fig.  16).     For  different  values  of  x^  t.hft 

function,  i/,  takes  different  values^  sgrpfitimps  inp.rpj^i^ing 

and  sometimes  decreasing.     Suppose  we  ask  the  questix)n: 

What  is  the  greatest  value  of  the  function  iar_the  range  of 

(It/ 
values  of  x  from  —  1  to  3?    By  (18)  p.  41,  -/  is  the  rate  of 


50 


CALCULUS  AND  GRAPHS 


change  of  y  with  respect  to  x,  and  by  pp.  34-35  (assuming 
that  X  continuously  increases)  the  function  y  increases  when 

-p  is  positive,  decreases  when  -^  is  negative.    The  function 
dx  (tx 

ceases  to  increase  and  begins  to  decrease  when  —  changes 
from  being  positive  to  being  negative;  that  is,  when  -^  is 

Q/X 

equal  to  zero.     Let  us  inquire  at  the  same  time,  what  is 

Y 


Y 

Fig.  16 

the  least  value  of  the  function  between  x  =  —  1  and  x  =  S? 

du 
Obviously  y  decreases  when  -^  is  negative:  increases  when 

ax 


dy 
dx 


is  positive;   ceases  to  decrease  and  begins  to  increase 

dv 
when  -^  changes  from  being  negative  to  being  positive,  or 

dv 
when  -r  =  0.*    In  either  case,  for  the  function  to  reach  a 
dx 

dv 
maximum  value  or  a  minimum  value,  must  ^^  =  0.    This, 

dx 

as  we  say,  is  a  necessary  condition  for  a  maximum  or 

*  Or  when  -7-  =  oo  .    This  case  will  not  be  considered. 
dx 


THE  DERIVATIVE 


51 


minimum  value  of  the  function.    But  it  is  not  a  sufficient 

dv 
condition:  that  is,  it  is  not  enough,  because  even  if  3^  =  0 

ax 

we  may  have  either  a  maximum  or  a  minimum  value  of 

the  function  y  or  may  have  neither  the  one  nor  the  other. 

Let  us  now  proceed  with  the  example.    Placing 

3^  = — h  3  a;  equal  to  zero, 

ax  2i 


^'  +  3  a:  =  3  X 


(-1)- 


we  obtain  x  =  0     and     x  =  2 

and  we  may  tabulate  our  information  as  follows: 


X 

—  1 

0 

1 

2 

3 

dy 
dx 

— 

0 

+ 

0 

— 

y 

Decreas- 
ing 

0 

Min. 

Value 

Increas- 
ing 

2 

Max. 
Value 

Decreas- 
ing 

That  is,  beginning  with  x  =  —  1  and  ending  with  a:  =  3, 
for  values  of  x  less  than  0,  y  decreases;  for  values  of  x  be- 
tween 0  and  2,  y  increases;  for  values  of  x  greater  than  2, 
y  decreases.  Therefore  the  least  value  of  the  function  (a 
minimum)  is  y  =  0,  when  x  =  0,  and  the  greatest  value 
(a  maximum)  is  y  =  2,  when  x  =  2. 

maximum 


It  should  be  noted  that  a 


mmimum 


value  is  not  nec- 


essai 


rily  the  j  ?  value  a  function  may  have.     For 


52 


CALCULUS  AND  GRAPHS 


instance  in  the  above  function,  x  =  —  3  gives  y  =  21 
which  is  greater  than  the  maximum  value  y  =  2,  and 
X  =  4:  gives  y  =  —  8  which  is  less  than  the  minimum  value 

^       .  I  maximum       ,      .  ,  .  ,    .    f  greater 

y  =  0-     ^{minimum  ^^^^^"^  ^«  °°^  ^^ich  lajfj^^*^'"  than 

the  values  immediately  before  and  immediately  after.  It 
will  be  seen  (Fig.  17)  that  at  a  maximum  or  minimum  point 


the  tangent  to  the  curve  is  parallel  to  the  axis  of  x,  so 

that  the  slope  is  zero;  that  is,  -^  =  0.    Just  before  a  maxi- 

dx 

mum  (point  P)  the  tangent  makes  an  acute  angle  with  0  X 

dv 
and  -J    =  ~^y  just  after  a  maximum  (point  Q)  the  tangent 

dv 
makes  an  obtuse  angle  with  0  X  and  -r  =  — •    Just  before 

dx 


8L  minimum  (R)  the  angle  is  obtuse  and  -r  = 

dx 


-;  just  after 


THE  DERIVATIVE 


63 


dv 
a  minimum  (S)  the  angle  is  acute  and  ;^  =  +.     These 

results  are  identical  with  those  obtained  from  the  point  of 
view  of  increasing  and  decreasing  values  of  the  function. 
We  may  summarize  the  results  as  follows : 

For  a  i     / .  value  of  y,  a  function  of  :r,  -7^  =  0  and 

minnnum  ax 

changes  frcnlitaf-asxinc^as... (22, 


Example  1.  Find  maximum  and  minimum  values  of  the 
function  2x^  —  3  a:'-  —  \2x  -\-  6. 


Put  ^  =  2  r^  —  3  a:'-  —  12  X  +  6 

6  X  —  12  =  6  (a:  +  1)  (a:  —  2)  =0 
1,2 


dU  y 

-r  =  i)  X- 
dx 


X 

—  2 

—  1 

0 

2 

3 

du 
dx 

+ 

0 

— 

0 

+ 

u 

incr. 

13 
Max. 

deer. 

—  14 
Min. 

incr. 

That  is,  the  maximum  value  of  the  function  is  13,  when 
X  =  —  1 ;  the  minimum  value  is  —  14,  when  x  =  2. 

Example  2.  Find  maximum  and  minimum  distances  from 
the  origin  of  a  point  which  moves  in  a  straight  line  accord- 
ing to  the  law  s  =  t^  —  12  <-  +  45  <  +  3  (feet;  seconds). 

♦  When  -7-    =  0  and  does  not  change  sign,  we  can  only  say  that  y  has  neither  a  maxi- 
dx 
mum  nor  a  minimum  value. 


54 


CALCULUS  AND  GRAPHS 


ds 
It 


=  3  <2  —  24  <  +  45  =  3  (« —  3)  (« —  5)  =  0 


whence  <  =  3  or  5. 


t 

0 

3 

4 

5 

6 

ds 
dt 

+ 

0 

— 

0 

+ 

s 

incr. 

57 

deer. 

53 

incr. 

that  is,  the  maximum  distance  is  57  feet  at  the  end  of  3 
seconds;  the  minimum,  53  feet  at  the  end  of  5  seconds. 
This  means  that  the  point,  starting  at  s  =  3  feet,  moves 
away  from  the  origin  until,  at  the  end  of  3  seconds,  it  is 
57  feet  away.  Then  it  moves  toward  the  origin  until,  at 
the  end  of  5  seconds,  it  is  53  feet  away.  After  that  it  moves 
away  from  the  origin  indefinitely.  The  maximum  s  is  not 
the  greatest  distance,  for  t  =  7,  for  example,  gives  s  =  73. 
The  minimum  s  is  not  the  least  distance,  ior  t  —  0  gives 
s  =  3. 


Example  3,  A  sheet  of  zinc  6  feet  long  and  3  feet  wide 
is  to  be  made  into  a  rectangular  tank  by  cutting  out  square 
corners  and  folding  up  the  edges.  Find  the  dimensions  of 
the  tank  when  its  capacity  is  a  maximum.    See  Fig. 

Let  rr  be  a  side  of  the  square 
cut  out  and  u  the  volume  of 
the  tank.  Then  u  =^  x  {%  —  2x) 
(3  —  2  a:) 


'V 

.^\ 

1 

1 
J 

6-2a; 

1 

l! 

-2sc 

u 

X 

-1 
1 
1 
1 

1 
1 

• 

6' 
Fig.  Ex.  3 


ovu  =  2{2x^  —  ^x'^  +  ^x) 


THE  DERIVATIVE 


55 


du 
dx 


=  2  (6  X-  —  18  a:  +  9)  =  0 


X  =  ^^y^  =  2.37  or  .64  feet 


The  first  value  is  obviously  impossible  as  it  is  more  than 
half  the  shorter  side  of  the  sheet  of  zinc.  Therefore  if 
there  be  a  maximum  value  it  will  occur  when  x  =  .64  feet. 
There  must  be  a  maximum  volume  since  when  a;  =  0  the 
volume  is  zero  and  increases  with  x.  When  x  =  1.5  feet 
the  volume  is  zero  and  increases  as  x  decreases.  The  volume 
increasing  from  both  extremes  must  reach  a  maximum  value 
In  most  practical  problems  it  is  obvious  that  there  must 
be  a  maximum  or  minimum,  as  the  case  may  be,  and  no 
further  test  need  be  applied.  In  the  above  problem,  there- 
fore, the  dimensions  for  a  maximum  are  length  4.72  feet* 
width  1.72  feet,  volume  5.2  cubic  feet. 

Example  4-  Find  maximum  and  minimum  velocities  of  a 
particle  moving  in  a  straight  line  according  to  the  law 
s  =  31^  —  2  f. 


at 


dv 
dt 


^=  18<-4  =  0,<  =  | 


t 

0 

2 

1 

dv 

dt 

— 

0 

+ 

V 

deer. 

Min. 

4 

incr. 

56 


CALCULUS  AND  GRAPHS 


^,   a  mathematical  minimum. 


Thus  <  =  I  gives  v  = 
This  is  not  the  least  speed,  however,  since  obviously, 
V  is  numerically  least  when  t  =  0.  There  is  no  maxi- 
mum. 


Example  5.  Find  the  dimensions  of  the  greatest  right 
circular  cylinder  inscriptible  in  a  right  circular  cone  of 

radius  r  and  height  h,  the 
planes  of  the  bases  of  the  cone 
and  cylinder  being  the  same. 
See  Fig. 

Let  V  be  the  volume  of  the 
cylinder.  .Then  v  =  tt  x^y 
which  is  a  function  of  two 
variables  x  and  y.  These  are 
not,  however,  independent, 
■^  since  the  triangles  ABC  and 
ADE  being  similar. 


AB       AD         h  —  y      h      ^  h  ,  . 


Therefore,  v  = x'^  (r  • 

r 


•  x)  =  —  {rx'^  —  x^) 
r 


dv  _  T  h 
dx         r 


(2  ro:  —  3  x")  =  0 


a;  =  0  or 


2r 


The  first  value  is  obviously  impossible,  and  the  sec- 
ond obviously  gives  a  maximum  cylinder  whose  dimen- 
sions are: 


2  r  h  .  TT  r^A       4      ,  « 

X  =  -;7-,  y  =  -^^f  V  —  4:    ^„     =  -  volume  oi  cone. 


3' 


27 


THE  DERIVATIVE 


57 


Example  G.  A  tin  tomato  can  is  to  be  made  in  the  form 
of  a  right  circular  cylinder.  What  will  be  its  dimensions 
when  the  amount  of  tin  used  is  least?    See  Fig. 

Let  u  be  the  total  surface  of  the  can.  

Then 

u  =  2Tr'^-{-2Trrh 

a  function  of  two  variables.  No  data  are 
given  by  which  we  can  connect  r  and  A,  but 
common  sense  tells  us  that  the  can  must  be 
of  a  definite  capacity,  so  that  we  may 
write  for  the  volume  of  the  can 


V  =  w  r-h  =  constant  =  a. 

From  this  the  value  of  h  might  be  expressed  in  terms  of  r 
and  substituted  in  Uj  as  in  Example  5;  or  we  may  proceed 
as  follows: 

$  =  27rf2r+/t+rf|   =  0  and 
dr  dr\ 


dr  [  dr\         dr 


From  the  latter  equation 


dh 
dr 


2h 
r 


which  being  substituted  in  the  former  gives 

2r+  h  —  2h  =  0 
or  h  =  2r, 

That  is,  for  a  cylindrical  can  of  any  definite  capacity,  the 
amount  of  material  will  be  least  when  the  height  is  equal 
to  the  diameter  of  the  base. 


58  CALCULUS  AND  GRAPHS 

EXAMPLES 

Find  maximum  and  minimum  values  of  the  following  functions. 
Sketch  the  graphs. 

1.  a;3+2x2                       2.  a;2  — 3x3  3,  ^^  ^  a;2  _|.  ^3 

4.  2x  —  x^  +x^               5.  3  —  2  x3  6.  4  a;  —  3  a;^ 

7.  sin  3x                          8.  sin  X  +  cos  x  9.  sin  x  —  cos  x 

In  the  following  examples  the  path  of  a  particle  moving  in  a  straight 
line  is  given,  s  being  distance  in  feet,  t  time  in  seconds.  In  each  case 
find  the  greatest  or  least  speed  of  the  particle  during  the  first  second. 

10.  s  =  St^  —  2t^+St  12.  s  =  16^  —  2^2  +  ^3 

11.  s  =^t  +  t^  —  t^ 

13.  Find  the  dimensions  of  the  rectangle  of  greatest  area  with  a 
perimeter  of  10  feet.  Draw  a  graph  showing  how  the  area  changes 
with  the  length  of  the  rectangle. 

14.  The  surface  of  a  hollow  cylinder  without  top  is  100  square  inches. 
Find  the  maximum  volume. 

15.  The  volume  of  a  solid  cylinder  is  V  cubic  inches.  Find  its  di- 
mensions if  the  total  surface  is  a  minimum. 

16.  A  running  track  is  in  the  form  of  a  rectangle  ABCD  with  semi- 
circles on  AB,  CD.  If  the  exterior  perimeter  is  a  quarter  of  a  mile, 
find  the  maximum  area  inclosed. 

17.  A  window  is  in  the  form  of  a  rectangle  surmounted  by  a  semi- 
circle. If  the  perimeter  is  30  feet,  find  the  dimensions  so  that  the 
greatest  possible  amount  of  light  may  be  admitted. 

18.  A  shot  is  projected  in  vacuo  with  velocity  u  feet  per  second  in 
a  direction  making  an  angle  a  with  the  horizontal.  Its  height  above  the 
ground  at  the  end  of  t  seconds  is  tu  sin  a  —  J  gt^.  Find  the  greatest 
height  and  the  time  of  reaching  it. 

19.  Find  the  circular  cylinder  of  largest  volume  which  can  be  cut  from 
a  sphere  of  radius  6  inches,  the  plane  ends  being  perpendicular  to  the  axis. 

20.  Given  200  square  feet  of  canvas,  find  the  greatest  conical  tent 
that  can  be  made  out  of  it. 

21.  Given  a  circular  sheet  of  paper,  find  the  angle  of  the  sector 
which  must  be  cut  out  so  that  the  remainder  may  be  folded  to  give  a 
conical  vessel  of  maximum  volume. 

22.  Find  the  least  area  of  canvas  that  can  be  used  to  construct  a 
conical  tent  whose  capacity  is  800  cubic  feet. 


THE  DERIVATIVE  59 

23.  Find  the  volume  of  the  greatest  right  cone  that  can  be  described 
by  the  revolution,  about  a  side,  of  a  right  triangle  of  hypotenuse  2  feet. 

24.  The  annual  cost  of  giving  a  certain  amount  of  electric  light  to  a 
town,  the  voltage  being  V  and  the  candle-power  of  each  lamp  C,  is 
found  to  be 

b 
A  =  a  +  --  for  electric  energy, 

and  ^      m       nV*^^  ^     , 

^  =  t;  +  ~7Jr~  ^or  1^"^P  renewals. 
C        C  '^ 

The  following  figures  are  known  when  C  is  10. 

V      100  200 

A     1500        1200 

5      300  500 

Find  a,h,  m,  n.  If  C  =  20,  what  value  of  V  will  give  the  minimum 
total  cost? 

25.  In  measuring  electric  current  by  a  tangent  galvanometer  the 
percentage  error  due  to  a  given  small  error  in  the  reading  is  proportional 
to  tan  X  +  cot  x.  Find  the  value  of  x  for  which  this  is  a  minimum. 
(Put  tan  X  =  t.) 

26.  An  electric  current  flows  around  a  coil  of  radius  a.  A  small  mag- 
net is  placed  with  its  axis  on  the  line  perpendicular  to  the  plane  of  the 
coil  through  its  center.  If  x  is  the  distance  of  the  magnet  from  the 
plane  of  the  coil,  the  force  exerted  on  it  by  the  current  is  proportional  to 


(x2  4-  a2)V2' 
Find  X  so  that  the  force  may  be  a  minimum.    (Put  x^  -\-  a^  =  y.) 

27.  A  ship,  P,  is  90  miles  south  of  another  ship,  Q.  P  is  sailing  north 
15  miles  per  hour,  and  Q  east  12  miles  per  hour.  How  long  will  the 
ships  approach  each  other?    When  will  they  be  nearest? 

28.  The  signals  between  two  boats  can  be  read  at  a  distance  of  one 
mile.  A  boat,  B,  sailing  due  south  at  6  knots  lies  5  miles  due  west  of 
a  boat,  C,  which  is  sailing  west  at  7  knots.  Do  the  boats  come  within 
signaling  distance?    When?    Why? 

29.  A  brace  in  the  form  of  the  letter  Y  is  to  be  16  feet  in  total  height 
and  12  feet  in  width  across  the  top.  Find  the  length  of  the  stem  and 
the  length  of  an  arm  if  the  length  of  the  stem  plus  that  of  the  two  equal 
arms  is  a  minimum. 


60  CALCULUS  AND  GRAPHS 

30.  A  rectangular  garden,  to  contain  4000  square  feet,  is  to  be  laid 
out  on  the  boundary  between  two  lots  and  to  be  fenced  in.  If  the  cost 
of  the  fence  along  the  boundary  is  to  be  shared  equally  by  the  two 
abuttors,  what  must  be  the  dimensions  of  the  garden  that  the  total 
cost  of  fencing  to  the  owner  of  the  garden  may  be  least? 

31.  A  window  in  the  form  of  a  rectangle  surmounted  by  an  equi- 
lateral triangle  is  35  feet  in  perimeter.  Find  its  dimensions  to  admit 
the  maximum  amount  of  light. 

32.  A  wire  10  feet  long  is  cut  into  two  pieces.  One  piece  is  made 
into  a  circle;  the  other  into  a  square.  Find  the  diameter  of  the  circle 
and  the  side  of  the  square  when  the  total  area  is  a  minimum. 

33.  A  tank  in  the  form  of  a  right  circular  cylinder,  open  at  the  top, 
is  to  hold  5000  cubic  feet.  The  bottom  costs  $2  per  square  foot;  the 
sides  $1.50  per  square  foot.    Find  the  dimensions  for  the  minimum  cost. 

34.  A  billboard  6  feet  high  stands  on  posts  10  feet  high.  How  far 
from  the  edge  of  the  road  must  the  board  be  placed  that  it  may  subtend 
the  greatest  angle  at  the  eye  of  a  person  5  feet  6  inches  tall?  (Sugges- 
tion: Find  maximum  value  of  tan  0,  when  0  is  the  subtending  angle.) 

16.  Differentials.    Approximate  Increments.    Errors.    If 

yhe  a  function  of  x  it  is  obvious  that 

A  2/  =  Al  .  A  X  (23) 

Z\  X 

even  when  A  y  and  A  x  are  small  numbers.     If  we  allow 

A  y 
A  a:  to  become  very  small  (to  approach  zero)  then  — -^  ap- 

dv 
proaches  -^^  as  a  limit.     To  indicate  that  A  x  and  A  y  are 
ax 

thus  becoming  very  small  we  shall  use  the  notation  dx  anjd 

dt/,  which  are  called  differentials,  and^hall  write 

dy^fj.  (24) 

That  is 

the  differential  of  y  equals  the  derivative  of  y  with 
respect  to  x  times  the  differential  of  x. 


THE  DERIVATIVE  61 

We  have   not  hitherto  used  -^^  as  a  fraction  but  (24) 

ax 

above  shows  that  if  we  use  it  as  a  fraction  and  cancel  the 
dx  the  result  is  the  differential  of  y  as  defined  above. 

It  is  obvious  that  the  increment  oi  y  (  \  y)  and  the  dif- 
ferential of  y  (dy)  are  not  equal,  but  it  is  also  obvious  that 
the  smaller  we  make  the  increment  of  a;  (Ax)  the  nearer 
will  A  y  and  dy  come  to  being  equal.  For  example  let 
y  =  x^j  then 

2 

A  2/  =  (^  +  A  x)"  —  x^  =  2x  ^  x  +   Ax 

^"*  2x  =  ^  =  ^,sothat 

dx       dx 

Ay  =  -^  Ax  +  Ax 

If  now  the  value  of  A  a:  be  taken  very  small,  the  value  of 

2 

A  X  becomes  very  small  even  as  compared  with  A  x;  *  so 
small  that  we  neglect  it  for  practical  purposes  and  write, 
when  Ax  or  dx  becomes  smaller  than  any  number  we 
please  to  assign,  the  differential  of  y^  in  general, 

and  for  the  function  y  =  x^ 

dy  =  2  xdx. 

We  thus  have  a  new  idea  (the  differential)  and  a  new  no- 
tation (the  differential  notation)  which  in  many  cases  are 
more  useful  and  convenient  than  the  derivative  and  the 
derivative  notation.  The  formulas,  (6)  p.  26  may  now  be 
written 

*  Let  A  a:  =  .00001,  for  example.    Then'A^^  =  ,0000000001. 


62  CALCULUS  AND  GRAPHS 

d(u  +  V  +  w  -{-  ...)  =  du+  dv  +  dw  -{-  ,  ,  , 

d(uv)  =  V  du-\-  udv 

dv!^  =  nu^~^  du 

d(c  u)  =  cdu 

dsinu  =  cos  u  du  (25) 

d  tan  u  =  sec^  u  du 

dsecu  =  sec  u  tan  u  du 

d  cos  u  =  —  sin  udu 

d  cot  u  =  —  csc^  u  du 

d  esc  u  =  —  CSC  u  cot  u  du 

It  should  be  noted  that  x  being  the  independent  variable 
we  can  give  its  increment,  A  x^  any  value  we  please  large 
or  small,  and  we  use  dx  merely  for  convenience  to  mean 
the  value  of  A  a:  when  it  approaches  zero.  We  can  not 
say  that  A  y  and  dy  are  the  same.  It  is  true,  however, 
that  in  the  case  of  continuous  functions,  the  kind  with 
which  we  deal,  A  y  approaches  zero  as  A  re  does,  and  in 
dealing  with  differentials  we  are  using  quantities  that  ap- 
proach zero  as  a  limit,  infinitesimal  quantities,  as  they  are 
called. 

Consider  the  equation  y  =  x^  which  may  be  used  to 
express  the  area  of  a  square  as  a  function  of  the  side.  Let 
X  =  3,  y  =  9,  Increase  x  by  .1;  then  y  -{-  Ay  =  (3.1)^  = 
9.61  and  Ay  =  9.61  —  9  =  .61.  If,  however,  we  increase 
A  re  by  .01,  we  find  Ay  =  .0601.  Let  us  write  the  differ- 
ential of  y, 

dy  =  2x  dx 

If  we  put  dx  =  .Ij  X  being  3,  we  have  dy  =  .6;  dx  =  .01, 
dy  =  .06.    Thus 


THE  DERIVATIVE 


63 


X 

Aa; 

Ay 

dy 

Ly  —  dy 

3 

.1 

.61 

.6 

.01 

3 

.01 

.0601 

.06 

.0001 

and  the  smaller  A  x  is  taken  the  nearer  will  the  values  of 
A  y  and  dy  be  to  each  other.  In  many  problems  the  error 
arising  from  the  use  of  dy  instead  of  A  i/  is  so  small  as  to 
be  negligible  to  the  degree  of  accuracy  with  which  we  are 
working.    For  example,  in  the  above  function  the  error  when 


dx  =  .1  is  A  7/  —  dy 


.01  or  II  =  1.6% 


When  dx  =  .01  the  error  is  A  i/  —  dy  =  .0001  or 


We  may  illustrate  the  difference 
between  Ay  and  dy  in  the  above  ^ 
function  graphically.  In  Fig.  18, 
ABGK  is  the  original  value  of  y; 
ACEH  the  new  value;  A  t/  is  3 
BCDG  +GDEF  +  GFHK.  But 
dy  is  BCDG  +  GFHK,  and  Ai/ 
—  dy  is  DEFG,  ^ 

Example  /.  Find  the  differential 
of  the  function  3  a:"*  +  \/x''  —  3. 


.0001 
.0601 


=  .17%. 


K  H 


Fia.  18 


Let  u  =  3  x^  +  Vx''  —  S  =  Sx'+  (x-  —  3)'/' 
Then 

d\3x'+  (x*^  — 3)V'f 


7         du     J 
du  =  T-  '  dx 
dx 


dx 


dx 


64  CALCULUS  AND  GRAPHS 

=  \l2x^+l^(x^  —  Sy'f'  ,2x\dx 

=  (l2x3+     /— )dx. 

\  Vx^  —  S/ 

Example  2.  Find  the  differential  of  sin^  3  x, 
d  (sin^  Sx)  =  2  sin  3  a;  d  (sin  3  x) 

=  2  sin  3  a;  .  cos  Sx  d  (Sx) 

=  2  sin  3  X  cos  Sx  .  S  dx 

=  6  sin  3  a;  cos  3  xdx  =  3  sin  6  xdx. 

Example  3.  A  circle  of  radius  3  feet  has  its  radius  in- 
creased by  .02  feet.  Find  approximately  the  increase  in 
area  of  the  circle. 

Here      A  =  wr^y 

dA  =  2  TT  rdr  and  when  x  =  S,  dr  =  .02, 
dA  =  (6  tt)  (.02)  =  .12  TT  square  feet. 

Example  4-  In  the  circle  of  example  3  what  is  the  per- 
centage error  made  in  using  dA  as  the  increase  in  area 
instead  of  A  A  the  actual  increase? 

A  A  =  TT  (3.02)2  —  w  (Sy  =  TT  (3.02^  —  3^)  =.1204  w 

square  feet. 

AA—dA  =  .1204  TT  —  .12  TT  =  .0004  tt  square  feet. 
AA—dA       .0004  TT 


A  A  .1204  TT 


=  .003  =  .3% 


Example  5,  Compute,  approximately,  the  volume  of  a 
cube  whose  edge  is  3.004  feet. 


THE  DERIVATIVE  65 

Let         V  =  x^j  then  dv  =  3  xHx. 
When     x  =  3  and  dx  =  .004,  then 

dv  =  (3)  (9)  (.004)  =  .108  cubic  feet. 
So  that  v'  =  v-\-  dv  =  27.108  cubic  feet. 

Example  6.  Find,  approximately,  the  value  of  sin  x  + 
2  cos  X  when  x  =  46°. 
Let        u  =  sin  X  +  2  cos  x  and,  when  x  =  45°, 

wi  =  sin  45°  +  2  cos  45°  =  ^-^  =  2.121. 

Li 

Also    di^  =  (cos  X  —  2  sin  x)  dx,  and  when  x  =  45% 
dx  =  1°  =  .017  radians, 

du  =  (cos  45°  — 2  sin  45°)  (.017)  =  _ -.^l^V^. 

Therefore  the  value  of  sin  x  -\-  2  cos  x  when  x  =  46°  is 

^^^3_^_.01^^2,983Vl^  2.109. 

id  Jj  4U 

EXAMPLES 

Find  the  differential  of  each  of  the  following  functions. 
1.  x»  — 3x2  2.  x  +  x2  — X*  3.  x3 -^  x'/=^  +  4    * 


' 

"•  -^     V^     ' 

5. 

a/x2-.2 

6.  (x2  +  1)'/' 

8. 

tan  5  X 

9.  sec2  \ 

11. 

cot  (3  X  +  1) 

12.  cos  (2  X  +  ^) 

4.  Vx^  +  ic 

7.  sin2  3x 

10.  X  +  sin  X 

Find  dy  from  each  of  the  following  functions: 

13.  X2/  =  20  14.  xhi  =  10  15.  xi/^  =  5 

16.  x2  +  a:?/  4-  2/2  =  2        17.  x^  +  2/2  =  16        18.  2  x^  —  3  2/2  =  6 

19.  Find  approximately  the  volume  of  a  thin  spherical  shell,  internal 
radius  1  foot,  thickness  .2  inch.  (Show  that  your  result  is  less  than  2 
per  cent  in  error.) 


66  CALCULUS  AND  GRAPHS 

20.  A  stone  is  thrown  at  an  angle  of  40°  to  the  horizontal  with  a 
velocity  of  80  feet  per  second.    Find  the  range  I  R  =  J  and 


/  V'  sin  2  d\ 

V  =  -r-) 


the  approximate  increase  in  the  range  if  the  velocity  is  changed  to 
82  feet  per  second,  the  angle  of  projection  remaining  unchanged. 

21.  A  cylindrical  well  is  said  to  be  25  feet  deep  and  6  feet  in  diameter. 
Find  the  error  in  the  calculated  volume  if  there  is  an  error  of  (1)  1  inch 
in  the  diameter,  (2)  3  inches  in  the  depth. 

22.  The  radius  of  the  base  of  a  cone  is  r  and  its  vertical  angle  2  a. 
Find  the  approximate  increase  in  volume  due  to  a  small  increase  A  r 
in  the  radius,  the  vertical  angle  remaining  constant.  Hence  show  that 
the  volume  of  a  conical  shell,  internal  radius  r,  thickness  t,  t  being 
small,  is  approximately  x  rlt  where  I  is  the  slant  height. 

23.  The  radius  of  a  sphere  is  found  by  measurement  to  be  18.5  inches, 
with  a  possible  error  of  .1  inch.  Find  the  consequent  errors  possible  in 
(1)  the  surface  area,  (2)  the  volume,  calculated  from  this  measurement. 

24.  The  side  of  a  square  is  measured  and  found  to  be  8  inches.  If 
an  error  of  0.01  inch  is  made  in  measuring  the  side,  find  approximately 
the  error  in  the  calculated  area. 

25.  If  the  edge  of  a  cube  is  measured  and  found  to  be  8  inches,  and 
if  an  error  of  V20  inch  has  been  made,  what  is  the  approximate  error 
in  the  calculated  volume? 

4 

26.  Find  approximately  the  value  of  7  x^  —  Sx  -\-  -  when  x  =  2.01. 

X 

27.  Find  approximately  the  value  of 

2  a:»  —  9  a;2  +  12  X  —  3  when  x  =  (1)  2.005,  (2)  1.995 
In  each  case  find  also  the  exact  value. 

28.  The  annual  cost  for  electric  energy  in  a  certain  plant  is  C  = 

a  +  --  where  V  is  the  voltage  and  a  and  b  are  constants.    It  is  observed 

that  a  =  900  and  h  =  60000.    If,  when  V  =  100,000  an  error  of  1000 
is  made  in  V  what  is  the  approximate  error  in  C? 

29.  A  right  triangle  by  measurement  has  sides  a  =  300  b  =  400  and 
c  =  500  feet  long  respectively.    Find  approximately 

(a)  The  error  in  the  area  of  the  triangle  if  there  is  an  error  of  5  inches 
in  measuring  o,  b  being  exact. 

(b)  The  error  in  the  area  for  an  error  of  5  inches  in  6,  a  being  exact. 

(c)  The  error  in  the  area  when  a  and  b  are  each  too  small  by  5  inches. 


THE  DERIVATIVE  67 

30.  A  right  triangle  has  legs  of  a  =  300  and  b  =  400  feet  in  length 
by  measurement.  Find  the  approximate  error  in  the  length  of  the 
hypothenuse  when 

(a)  The  legs  have  each  a  positive  error  of  one  per  cent. 

(b)  Each  a  negative  error  of  one  per  cent. 

(c)  Leg  a  a  positive  and  b  a  negative  error  of  one  per  cent. 

31.  The  resistance  of  a  circuit  was  found  by  using  the  formula 
C  =  E/Rj  where  C  =  current  and  E  =  electromotive  force.  If  there 
is  an  error  of  Vio  ampere  in  reading  C  and  V20  volt  in  reading  E,  what  is 
the  error  in  R  if  readings  are  C  =  20  amperes  and  E  =  120  volts? 
\Vhat  is  the  maximum  percentage  error? 

32.  If  the  formula  sin  (x  +  y)  =  sin  xcosy  -\-  cos  x  sin  y  were  used 
to  calculate  sin  {x  -{-  y),  what  approximate  error  would  result  if  an 
error  of  0.1  were  made  in  measuring  both  x  and  1/,  the  measurements 
of  the  two  acute  angles  giving  sin  x  =  ^U  and  sin  y  =  ^/u? 

33.  The  acceleration  of  a  particle  down  an  inclined  plane  is  given 
by  /  =  ^  sin  a.  If  a,  which  is  measured  as  30°,  may  be  in  error  .01°, 
what  is  the  maximum  error  in  ff    Take  ^  =  32  ft.  sec^. 

34.  The  period  of  a  pendulum  is  P  =  2  X4  /-  .    What  is  the  great- 

est  error  in  the  period  if  there  is  an  error  of  =fc  ^  foot  in  measuring  a 
10-foot  suspension?  What  is  the  percentage  error?   {g  equals  32  ft.  sec^) . 

35.  The  length  L  and  the  period  P  are  connected  by  the  equation 
4  x2  L  =  P^g.  If  L  is  calculated  assuming  P  =  1,  and  ^  =  32  ft.  sec^, 
what  is  the  approximate  error  in  L  if  the  true  value  is  P  =  1.02?  What 
is  the  percentage  error? 

36.  A  particle  is  moving  on  the  ellipse  x^  +  4  1/^  =  20  and  is  at  the 
point  (2,  2).  What  is  the  approximate  change  in  y  due  to  a  change  of 
0.1  in  x? 

37.  The  force  of  attraction  between  two  equal  masses  is  P  =  — , 

s- 

where  m  is  the  value  of  each  mass  and  s  the  distance  between  them. 
If  m  is  measured  as  10  grammes  and  s  as  5  centimeters,  find  the  error 
in  F  when 

(a)  8  is  exact  and  dm  =  =*=  .01  gr. 

(b)  m  is  exact  and  (is  =  =*=  .01  cm. 

(c)  dm  =  .01  gr.  and  ds  =  .01  cm. 

(d)  dm  =  .01  gr.  and  ds  =  —  .01  cm. 


CHAPTER  V 

INTEGRATION.     INDEFINITE  INTEGRALS. 
METHODS  OF  INTEGRATION 

17.  Integration.     The  Indefinite  Integral.     In  all  our 

work  hitherto  we  have  had  given,  y,  a  function  of  an  inde- 
pendent variable  x,  and  have  asked  what  is  the  derivative 
(or  differential)  of  y  with  respect  to  x.  We  have  also  learned 
how  to  obtain  the  derivative.  Let  us  now  assume  that  the 
derivative  is  given  and  ask  what  is  the  function  of  which 
it  is  the  derivative?    For  example  if  we  have  the  function 

y  =  x'^  we  know  that  -^  =  2x  or  that  dy  =  2x  dx,^     It 

is  obvious  that  we  can  write 

The  function  of  which  2  x  dx  is  the  differential  is  x^  simply 
by  using  inversely,  backwards,  so  to  speak,  our  knowledge 
of  the  result  of  differentiation. 

Our  next  thought  would  be  to  devise  some  simple  way 
of  making  our  statement  in  symbols  instead  of  words,  and 
the  notation  used  is  as  follows: 


p 


2x  dx  =  x^. 

Thus    j  2xdx  =  x'^  states  in  symbols  that  the  function  of 

which  2xdx  is  the  differential  is  x^;  more  exactly  that  it 
may  he  x^.  It  might  equally  well  he  x'^  -{-  2  or  x'^  —  5  or 
x'^  +  any  constant  whatever,  since  the  derivative  of  a;^  +  c 
is  2  X.    The  complete  statement  is,  therefore. 


/ 


2xdx  =  x'^-\-  c  (26) 

*  In  this  chapter  we  shall  use  differentials  and  the  differential  notation  as  being 
more  convenient. 


INTEGRATION  69 

Having  introduced  this  new  symbol  we  give  a  name  to  it, 
namely  integral,  and  the  expression  (26)  is  read: 

The  integral  of  2x  dx  (or  the  integral  of  2  x  with  re- 
spect to  x)  is  x^  +  c.  The  process  of  finding  the  integral 
is  called  integration. 

18.  Methods  of  Integration.  Recognition,  Transforma- 
tion, Substitution.  There  are  various  methods  of  integra- 
tion, of  which  we  shall  use  here  only  three,*  namely: 

I.  Recognition.    11.  Transformation.    III.  Substitution. 

The  first  method  we  have  already  used  in  the  above 
example,  j  2 x  dx  =  x'  +  c.  It  consists  simply  in  recog- 
nizing the  expression  to  be  integrated,  the  integrand  as  it 
is  called,  as  the  differential  of  a  certain  function.  The 
recognition,  however,  can  be  generalized  and  the  process 
simplified  by  introducing  certain  theorems  or  formulas  of 
integration.    Thus  we  know  that 

d  {u  -jr  V  -\r  w  -jr  '  '  ')  =  du  -\-  dv  -\-  dw  -\-  .  .  , 

But  the  function  of  which  d  (u  -\-  v  +  w  -}-,.,)  is  the  dif- 
ferential is  obviously  ^i  -\-  v  -\-  w  -]-,,.  y  and  this  is  the 
function  of  which  du  is  the  differential  plus  the  function  of 
which  dv  is  the  differential,  etc.    In  symbols  f 

j  {du  +  dv  +  dw+  ...)  =    fdu  +  J  dv  +    fdw+.  .  .(27) 

Stated  briefly  in  words,  ^*  the  integral  of  a  sum  equals  the 

sum  of  the  integrals  of  the  terms.'' 

In  a  similar  way  it  may  be  shown  that,  a  being  a  constant, 

j  a  du  =  a  J  du  (28) 

*  A  fourth  method,  integration  by  parts,  ia  explained  in  Art.  34  following. 

fTho  ron.stant  of  integration  will,  in  general,  be  oniitt«Ki.     It  is  always  understood. 


70  CALCULUS  AND  GRAPHS 

In  words,  "  a  constant  factor  can  be  put  outside  of  the  in- 
tegral sign." 

Again  we  know  that 

dx^  =  n  x^~^  dx 
Therefore, 

or  by  (28) 


fnx''-^  dx  =  Jdx^ 

njx''-^dx  =  aj" 
J  n 


s- 


which  can  be  put  in  a  rather  better  form  by  writing  n  —  1 
=  m  and  n  =  m  +  1,  thus 

x^  dx  =  ^ (29) 

m  +  1 

The  above  formulas  are  grouped  here  for  convenience  of 
reference,  together  with  four  others  immediately  recog- 
nizable. 

i.    I  {du  +  dv+  dw+  . .  .)=  j  du+  j  dv  +  j  dw-}-  , ,  . 

ii.    j  adu  =  a  I  du  (30) 

J  m+1 

iv.    j  sin  udu  =  —  cos  u  v.    j  cos  udu  =  sin  u 

vi.    f  sec^ udu  =  tan  i^  vii.    I  csc^ udu=  —  cot u 

The  second  method  of  integration  is  called  transformation. 
It  consists  merely  in  changing  the  form  of  the  integrand, 

*  This  formula  fails  when  m  =  —  1. 
t  This  formula  fails  when  m  =  —  1. 


INTEGRATION  71 

the  expression  to  be  integrated,  so  that  we  can  make  use 
of  the  first  method;  can  recognize  the  integrand  as  coming 
under  one  or  more  of  the  forms  of  (30).  No  rules  can  be 
given  for  the  transformation  to  be  employed,  as  we  can 
use  any  algebraic  or  trigonometric  transformation  that  may 
seem  available.  One  or  two  examples  will  illustrate  the 
process.  It  should  be  borne  in  mind  that  the  object  of  the 
transformation  is  to  obtain  one  or  more  forms  that  can  be 
recognized. 

Example  1.      rx^  +  Sx+2,  ,         ... 

I 7= ax  may  be  written 

•^  Vx 

/x^  Cxdx  rdx         ^ 

^  d.  +  Sj  -^-  +  2j  ;^-  by  (30)-x  and  n 


/ x''  dx  +  S  jx''  dx+2  jx''^'  dx  which  equals  by  (30)— ii 


x''-'  ,    3^      2^'-      2 
«  ^         5 


ii  S  1  " 

2  2^ 

The  result  of  integration  can  always  be  tested  by  differen- 
tiation. 

Thus  using  the  result  of  Example  1  we  have: 

d{ix'''+2x'^  +  4x''')  _2     5    V  3    V  1     ./= 

dx  -  5-2''    ^''-  2""    ^*-2'' 

X'' 


Vx 


Example  2.      T  •  9  ^   ,  n 


2 

We  know  that     2  sin- 17  =  1  —  cos  6 

or  .  ,  e       1       1        a 

"'""2  =2~2'°'^ 


72  CALCULUS  AND  GRAPHS 

Therefore 

—  ^  Jcos  ddO      (30)— i  and  ii 

=  ^0-|sin0     (30)-v 

Example  3, 
fV^  (x  —  2)2  dx  =    fx''  (x'-  —  4:x  +  4)dx  = 

C{x^'—4x^'  +  4 xf')  dx  =  fxl' dx—^  fx''' dx  +  4:  Cx^' dx 

X''       4:xl'  .   4:x''       2    7/,       8    6/,   ,    8    3/, 
=  —^ : 1 =  -  x'^ x"  +  -  x'\ 


^7  5  3 


2^  2^  2^ 

The  third  method  of  integration  is  called  substitution  of 
a  new  variable.  It  consists  in  choosing  a  new  variable, 
related  to  the  original  variable  in  some  certain  way,  so 
that  when  we  substitute  and  obtain  the  integrand  in  terms 
of  the  new  variable  we  can  recognize  it  by  the  first  method. 
Every  method  of  integration  in  the  end  reduces  to  the 
recognition  of  fundamental  forms  such  as  those  given  in 
(30). 


Example 


'■  JV7- 


-\-  I  .  X  dx 


du 
Put  a;^  +  1  =  u;  then  2x  dx  =  du,  xdx  =  —  and  we  may 

write 


INTEGRATION  73 


Or   we   may   put    Va;-  +  1  =  u,   x^  +  1  =  u'^,     2xdx  = 
2  u  du,  xdx  =  u  du.    Then 

fVx'^  +  1  xdx  =  1'^  '  '^d'^  =  lu^du==  ^  =  1  (a;2  +  !)•/« 

the  same  result  as  above. 

Example  2,      /  sin  2  a;  dx. 

du 
Put  2  X  =  u,  2  dx  =  du,  dx  =  — .  Then 


j  iim2  x  dx  =  I  sin  u  .  —  =  -  j  sin  udu  == 


du     1  r  ^      ,  1 

-  cos  u 


=  —  -  COS  2  x. 

There  are  many  rules  for  the  choice  of  a  new  variable  for 
special  forms  of  functions.  We  shall  give  only  one  general 
rule. 

If  one  part  of  the  integrand  is  (except,  perhaps,  for  a 
constant  factor)  the  differential  of  another  part,  put  that 
other  part  equal  to  a  new  variable  and  substitute.  (31) 

In  general  it  may  be  said  if  recognition  and  transforma- 
tion fail,  try  some  substitution. 

Example  1 .      fix''  +  5)  x''  dx. 

Here  x^^  dx  is  the  differential  (constant  factor  excepted)  of 
x'-  +  5,  therefore,  by  (31)  we  put 

x'''  +5  =  1/,   -  x'^  dx  —  du,  x'""  dx  =  -  du.  Then 

I  {x"  +  5)  x"  dx  =  \u.-du  =  -\udu  =  -.— 


74                     CALCULUS  AND  GRAPHS  I 

Or  we  might  proceed  thus:  i 

Jix'l'  +  5)  xl' dx  =f(x^  +  5 x'')  dx  =  JxHx  +  5  fa;'/' dx  ■ ' 

^t  +  5    ^'_^      10  ,/,  I 

The  results  by  the  two  methods  do  not  seem  to  agree,  but  | 
if  we  expand  the  former  we  have 

and  the  results  differ  only  by  the  constant  -^/.    Since  the  I 
constant  of  integration,  though  not  written,  is  understood 
in  every  case,  both  results  are  correct.     Two  results  that 

differ  by  a  constant  term  are  both  correct.  j 

i 

Example  2,                 I  sin  x  cos  xdx  ■ 

Put                   cos  X  =  Uy     —  sin  a;  da:  =  du;                  then  i 

/,              C     J             y^           cos^o; 
sm  X  cos  xdx  =  —  I   udu  =  —  —  =  — 
J                      2                2 

or,  put  sin  x  =  u,      cos  xdx  =  du 

/J         r    7         ^^      sin^  X  \ 

sin  X  cos  xdx  =  j  udu  =  —  =  — ——  ] 

But  sin^  X  =  1  —  cos^  x;  therefore 

sin  X       \       \  ■ 

— -—  =  -  —  -  cos^  X  which  differs  from  the  first  result  by  j 

2          2       2  I 

the  constant  |^.     Or,   again,   since  2  sin  x  cos  a;  =  sin  2  re  j 

I  sin  X  cos  xdx  =  -  j  sin  2  xdx  ] 


INTEGRATION  75 

Put  2x  =  u^    2  dx  =  dUj    dx  =  -  du         and 

2 

I  sin  X  cos  a;  da:  =  -  /  sin  2  a;  da;  =  -  I  sin  udu 

=  —  -  cos  u  =  —  7  cos  2  a;  =  —  -  (cos^  x  —  sin^  x) 
4  4  4 

=  —  -  (cos^  X  —  1  +  cos^  x)  =  —  -  cos^  ^  "I"  I 

The  three  forms  of  the  result  are  equally  correct.  Also  it 
will  be  noted  (last  method)  that  we  sometimes  use  both  a 
transformation  and  a  substitution  before  we  recognize  the 
integral. 

Example  3.  Find  the  function,  y^  of  which  a;^  +  3  a;  is 
the  derivative,  given  that  y  =  6  when  a:  =  1. 

y  =f(x^^  +  Sx)  dx  =  "^  +  ^  +  c 

But  y  =  6y  X  =  Ij  therefore 

^1,3,  25  , 

32'6 

^x\     Sx^      25 

^       3         2    "*"  6 

Example  4-  The  slope  of  a  curve  which  passes  through 

dv      X 
X  =  I   y  =  S  is  -r  =  -,    Find  the  equation  of  the  curve. 
dx      y 

We  have 


dy      X         ,           J 
-r-  =  -  or  y  dy  =  X  dx 
dx       y       -^    "^ 

therefore 

j  ydy  =   j  xdx 

or 

o  =  ";7  +  c 

76  CALCULUS  AND  GRAPHS 

9       1 
But  X  =  1,  y  =  S;  therefore,  -=-  +  corc  =  4  and  the 

equation  is 

u"  X 

-  =  -  +  4  or  t/2  —  0^2  =  8. 

Example  5,  The  velocity  at  any  time,  t,  of  a  particle  mov- 
ing in  a  straight  line  is  z;  =  5  —  3  ^.  Express  the  distance 
as  a  function  of  the  time,  given  that  s  =  15  when  t  =  2, 

We  have  ds       _       ^ ,        ,        ,.       o  a  ^/ 

-—  =  5  —  3  ^  or  as  =  (5  —  6i)  at 
at 

Therefore  A.       o  ^\  ^^       r  ^       ^  ^^   i 

s  =  J  (5  —  St)  dt  =  5t —  +  c 

But  s  =  15,  ^  =  2,  therefore  15  =  10  —  6  +  c,  c  =  11  and 
the  equation  is 

It  should  be  noted,  in  connection  with  this  problem,  that 
the  constant  of  integration  (c  =  11)  denotes  the  position  of 
the  particle  when  it  begins  to  move;  that  is,  the  value  of 
s  when  t  =  0. 

EXAMPLES 

Integrate  the  following  functions  and  prove  your  result  correct. 

1.  J  (1  +  Vx)  dx     2.  J  iy/x-k-  x^)  dx      3.  J  f  ^  —  77^  V^ 

4.  J  (a;V3  —  x-'l^)  dx    5.  j{x  +  ajV4)  dx         6.  J  ( Vi+ "V^x) Jx 

Integrate  each  of  the  following  functions  by  two  methods  and  show 
that  the  results  differ  only  by  the  constant  of  integration. 

7.    j(l—xydx  8.     /(2+x)2dx      .     9.    jix  —  Sydx 

10.    jiu+^ydu        11.     /(s  — 5)2ds  12.    J  (0—^/2yd^ 


INTEGRATION  77 

Integrate  the  following  functions; 

J\        x'7  J  (5  — s*)8 

15.     /(l  — -)(te  30.     I- -— 

f/     i\2 ,  „.  r    s'^'^ 

17.  J (y"-v-'Vdy  32.  J^-^^, 

18./(V;+A^Q^<fe  33. /^-^J^ 

j{S  —  5x)-'dx  34.    I  {x  +  l)Vx^+2xdx 

/5dx  ^^      r  (x  +  1)  dx 
(1+2  a;)2  *  J  -v/x2  +  2  X 

/  \/4  —  3  ^  ctt  36.     /  \/3  x2  —  6  aj  .  (a;  —  ] 


19 


20 


21.  IV4:  —  3tdt  36.     I  V3x2  — 6aj  .  (a;  — l)cte 

22.  I  \/2  w  +  1  (iw  37.     I  sin2  ^  cos  -  dx 

/d  0  ^^      /^cos  2  w  rfi 

.  38.     I  -^— - — 

\/l  4-4  0  •^^    sm2  2  w 

39.     r  sin  tdt 


r     dt  39.  r 

./  V2  — 3«  *^  Vcos« 

25.     I  X  (x2  +  1)3  (te  40.     /  cos'  3  x  sin  3  a;  dx 


5.    I  X  (x2  +  1)3  (te  40.     /  ( 


(ix 
3x  dx 


26. 

27.     I  w2  (3  +  i^')*  du  42.     I  cos  4  x  sin  4  x  dx 


78  CALCULUS  AND  GRAPHS 

In  the  following  examples  find  the  function,  2/,  of  which  the  given 
expression  is  the  derivative  under  the  conditions  named. 

4S.  2x^  —  x;y  =  5,x  =  0  47.  5  ^  +  4;  2/  =  3,  ^  =  3 

44.  x  —  x\'y  =  2,x  =  l  48.  4  —  5  <;  2/  =  4,  «  =  2 

45.  X  +  \/x;  y  =  0,x  =4:  49.  32  ^  +  6;  ?/  =  5,  i  =  1 

2/  =  16,  ^  =  2 

46.  x2  +  a:  +  2;  2/  =  1,  X  =  1  50.  32  t  —  b;  y  =  4,  t  =  1 

y  =  20,t  =3 

5L  The  slope  of  a  curve  is  given  by  —  =  3  +  4  a;.    Find  the  equa- 

dx 

tion  of  the  curve  if  it  passes  through  the  point  x  =  1,  y  =2. 

52.  The  slope  of  a  curve  is  given  by  -7-  =  2x  —  5.    Find  the  equa- 

dx 

tion  of  the  curve  if  it  passes  through  the  point  x  =  2,  y  =2. 

53.  A  body  moves  in  a  straight  line  so  that  its  velocity  at  any  time 
t  is  given  by  v  =  16  —  41.  Express  the  distance,  s,  as  a  function  of 
t,  given  that  s  =  20  when  ^  =  5. 

54.  A  body  moves  in  a  straight  line  so  that  its  velocity  at  any  time 
t  is  given  by  z;  =  16  +  3  ^.  Express  the  distance,  s,  as  a  function  of 
t,  given  that  s  =  100  when  t  =  2. 

55.  A  body  moves  in  a  straight  line  so  that  its  velocity  at  any  time 
t  is  given  by  y  =  16  +  ht.  Express  the  distance,  s,  as  a  function  of  t, 
given  that  s  =  10  when  t  =  1  and  s  =  100  when  t  =  5. 

56.  A  body  moves  in  a  straight  line  so  that  its  velocity  at  any  time 
t  is  given  by  y  =  16  —  bt.  Express  the  distance,  s,  as  a  function  of  t, 
given  that  s  =  0  when  t  =  1  and  s  =  20  when  t  =  3. 

57.  A  particle  rolls  down  a  smooth  inclined  plane  so  that  v  =  14.2  t. 
Express  the  distance  it  rolls  (s)  as  a  function  of  t.  How  far  will 
it  roll  in  the  time  between  t  =  0  and  t  =  2?  In  the  second  and 
third  seconds,  the  unit  of  time  being  the  second,  the  unit  of  distance 
the  foot? 

Find  the  function  of  which  each  of  the  following  expressions  is  the 
derivative,  and  sketch  the  graph  of  the  function  when  the  value  of  the 
constant  of  integration  isc  =0,  c  =  l,c  =  — 2. 


INTEGRATION  79 


58.  3 

61,  2x 

59.  3a;  — 2 

62.  a;«  — 2x+3 

60.  3  — X  — x« 

63.  4a;» 

64.  t;  =  4 

67.  t;  =  4  — « 

65.  t;  =  2 «  +  3 

68.  t;  =  3  <« 

66.  V  =3i»+2^- 

-8 

69..  =  ^, 

In  examples  64-69 

vis 

velocity,  i 

is  time. 

CHAPTER  VI 

INTEGRATION.    APPROXIMATE  SUMMATION.    THE 
DEFINITE  INTEGRAL 


19,  Approximate  Summation.    Let  us  consider  the  func- 
tion y  =  x^,  of  which  the  graph  is  given  in  Fig.  19,  and  ask, 

7f« 


what  is  the  area  bounded  by  the  curve,  the  axis  of  x  and 
the  two  ordinates  corresponding  to  rr  =  1  and  a:  =  4;  in 
the  figure  the  area  of  PMSQf  To  obtain  this  area  we 
proceed  in  a  similar  manner  to  that  used  in  the  elementary 
geometry  to  obtain  the  area  of  a  circle.  We  divide  the 
line  MS  into  any  convenient  number  of  parts — three  in  the 

80 


INTEGRATION 


81 


figure — and  at  the  points  of  division  erect  ordinates  to  the 
curve,  MPy  NA^  RB  and  SQ.  At  the  points  where  these 
ordinates  touch  the  curve  we  draw  PC^  AD  and  BE  par- 
allel to  the  axis  of  x.  We  thus  have  three  rectangles  con- 
structed, inscribed  in  the  curve,  and  can  compute  the  areas 
of  the  rectangles.    Thus 

PMNC  =  MPX  MN  =  1X1  =  1 
ANRD  =  NAX  NR  =  4.Xl  =  4: 
BRSE  =  RBXRS    =9X1  =  9 
and  the  sum  of  the  rectangles  equals   1  +  4  +  9  =  14, 
which  is  obviously  smaller  than  the  area  bounded  by  the 
curve  by  the  sum  of  the  triangular  pieces  PC  A,  ADB^ 


iy2  2   2^2 

Fig.  20 

etc.,  left  over.  Let  us  next  double  the  number  of  rectangles 
by  bisecting  each  of  the  lines  MN,  NR,  etc.,  and  con- 
structing the  new  set  of  inscribed  rectangles,  as  shown  in 
Fig.  20.    The  area  would  now  be,  since 


82  CALCULUS  AND  GRAPHS 

n  =  area  PA  =  MP  X  MA  =  IX  k  =  ^ 

r2  =  area  BiV  =  ^fi  X  A  iV  =  (1  +  1)2  X  I  =  I  * 

n  =  (2)2  X  I  =  2 

U=  i2  +  irxk  =  -%'- 

n  =  (3)2  X  i  =  I 

»-6=  (3  +  1)2X1  =  -*#     . 

and  the  sum  of  the  rectangles  is 

Sr  +=  ^  +  1+  2  +  ¥  +  I  +  -V-  =  ^1^  =  17|  =  17.4 

Obviously  this  result  is  nearer  than  the  first  result  (14 
square  units)  to  the  area  of  the  curve,  as  smaller  triangular 
pieces  have  been  neglected.  By  making  the  number  of 
rectangles  greater  and  greater  we  can  make  the  sum  of 
the  areas  of  the  rectangles  nearer  and  nearer  to  the  area 
of  the  curve.  By  taking  the  number  of  rectangles  great 
enough  we  can  make  the  difference  between  the  area  of 
the  curve  and  the  area-sum  of  the  rectangles  less  than  any 
assigned  value,  however  small.  In  other  words,  the  area 
bounded  by  the  curve,  two  ordinates  and  the  axis  of  x  is 
the  limit  of  the  sum  of  the  inscribed  rectangles  as  the 
number  of  rectangles  is  indefinitely  increased. J    In  symbols 

Curve-area  =  limit  Sr  (32) 

We  do  not  yet  know  how  to  get  the  limit  of  this  sum  and 
for  the  present  shall  content  ourselves  with  obtaining 
closer  and  closer  numerical  approximations. 

It  is  evident  that  we  might  have  circumscribed  rectangles 
(see  Fig.  19)  instead  of  inscribing  them,  thus  getting  suc- 
cessive  approximations   always   a   little   larger   than   the 

*  AB  =  (1  -f  H)2,  since  the  curve  is  y  =  x^  and  the  ordinate  is  therefore  always 
the  square  of  the  corresponding  value  of  the  abscissa. 
t  Read,  Sigma  R  and  meaning  the  sum  of  the  r's. 
X  Compare  this  discussion  with  that  in  Articles  5  and  following,  Chap.  II. 


INTEGRATION 


83 


curve-area.    In  the  first  case  above  we  should  have  ob- 
tained, using  R  to  mean  any  circumscribed  rectangle, 

fii  =  4  X  1  =  4,  ffj  =  9  X  1  =  9,  i23  =  16  X  1  =  16  and 

Si?  =  4  -h  9  +  16  =  29 
We  have  already  seen  that  in  this  case 

2r  =  14 

so  that  the  curve-area  lies  between  14  and  29  square  units, 
a  very  rough  approximation. 

Using  the  second  series  of  circumscribed  rectangles, 
where  we  pass  along  OX  at  intervals  of  one-half  (that  is, 
A  a;  =  ^)  we  have 

fii  =  (1  +  ^r  a)  =  I    1       and 

R2  =  (2)^  a)  =  2 

«3  =  (2  +  i)M^)  =?^ 

R*  =  (3)^  (i)  =  I 

ft5  =  (3  -t-  hV  (1)  =  ^ 

Ke  =  (4)^  a)  =  8 

We  have  already  seen  that  in  this  case 

Sr  =  17|  =  17.4 

so  that  the  curve-area  lies  between  17.4  and  24.9  square 
units,  a  closer  approximation. 

A  closer  approximation  still,  and  more  quickly  obtained, 
is  got  by  taking  the  half  sum  of  the  inscribed  and  circum- 
scribed rectangles.  This  is  equivalent  to  drawing  straight 
lines  PA,  AB,  BQ,  etc.  (Fig.  19)  and  taking  the  sum  of 
the  trapezoids  APMN,  etc.    Thus  in  the  above  problem: 

Case  1.  Curve-area  =  ^(  2r  -|-  Si?)  =  ^(14  +  29)  =  21.5 

Case  2.  Curve-ai-ea  =  ^  (  2r  -f  2ft)  =  ^  (17.4  +  24.9)  = 
21.1 


2ft  =  |-f  2  +  -^/  +  |  +  -*^  +  8 
=  i|iL  =  241  =  24.9 


84 


CALCULUS  AND  GRAPHS 


We  shall  find  later  that  the  exact  area  in  this  case  is 

Curve-area  =  UmitSr  =  limit  Si2  =  21. 

The  subject  of  approximate  summation  is  so  important 
that  we  shall  work  another  illustrative  example. 

Example.  Find  approximately  the  area  lying  above  0  X 
bounded  by  the  curve  y  =  \/x,  the  axis  of  x  and  the  two 
ordinates  corresponding  to  x  =  |^  and  x  =  4,  taking  inter- 
vals along  0  X  oi  /\  X  =  ^.    Fig.  2L 


Q 


B 


M 


D 


N 


2 

Fig.  21 


We  have,  proceeding  as  in  the  first  example,  since  y  =  \/x, 

AB  =  Vi,  MC  =  AP  =  \/l,  etc.,  and  OA  =  AM  =  etc. 

=  Ax   =  |. 


2r  =  iVi)  (I)  +_(Vl)  (i)  +  (Vl_+  i)  (i)  +(V^  (^) 
+  (\/2  +  i)  (i)  +  ( V3)  (^)  +  (VS  +  ^)  (^) 


INTEGRATION  85 


and 


sft = (Vi)  (h) + (>/]_+  i)  a) + (V2)  a) +(a/2+ ^)  a) 
+  ( V3)  a) + ( V3  + 1)  (1)  +  (v*)  a) 


=  ^1  Vl  +  Vl+i  +  \/2  +  \/2  +  i  +  a/3 
+  A/3  +  i+V4l 
Then 


Sr  +  2 ft  =  1 1  A/i  +  2A/i+  2VlJhj_+  2a/2_+  2a/2  +  ^ 
+  2V3+2a/3  +  1  +  \/4[ 

=  ^1 .71  +  2  +  2.44  +  2.82  +  3.16  +  3.46 
+  3.74  +  2  i 

=  ^|  20.33  [ 
.and 

curve-area  =  i  (  Sr  +  Sft)  =\  (20.33)  =  5.08 

The  exact  area  to  the  second  decimal  place  in  this  case,  as 
will  he  seen  later,  is 

Curve-area  =  limit  Sr  =  limit  S/?  =  ^  — ^  =  5.09 

EXAMPLES 

In  the  following  examples  sketch  the  graph  and  find  the  approxi- 
mate curve-area  between  the  limits  named  at  the  intervals  A  x  =  .5 
and  A  a;  =  .2. 

1.  !/  =  x2  +  2.  From  a;  =  0  to  x  =  3. 

2.  y  =  x^  —  9.  From  x  =  0  to  x  =  2. 

3.  2/  =  4  —  x2.  From  x=— 2tox=+2. 

4.  2/  =  9  —  x2.  From  x=— 3tox=-i-3. 


86 


CALCULUS  AND  GRAPHS 


5.  4  2/  =  3  +  2  X  —  a;2. 

6.  4  2/  =  3  +  2  x  —  x2. 

7.  9  2/  =  5  +  4  X  —  x2. 

8.  9  2/  =  5  +  4  a;  —  x2. 

9.  2/  =  2  a;3. 

10.  2/  =  3  a;3. 

11.  2/  =  sin  X. 

12.  2/  =  sin  a;. 

13.  2/  =  cos  X 

14.  2/  =  cos  X. 


From  a; 
From  X 
From  X 


0  to  X  =  2. 
—  1  to  X  =  3. 
0  to  X  =  2. 


From  X  =  —  1  to  x  =  5. 

From  X  =  1  to  X  =  3. 

From  X  =  1  to  X  =  3. 

J^rom  X  =  ^  to  X  =  -4-;  A  x  =  g^. 

From  X  =  0  to  X  =  x;  A  x  =  § . 

From  x  =  —  4tox  =  f;Ax=8. 

From  X 


5 
12| 


6      7        8       9 
14    11^    15     14 


0  to  X  =  x;  A  X  =  8^. 

15.  The  depth  of  the  water  on  the  face  of  a  vertical  dam  is  measured 
at  intervals  of  one  foot  and  found  to  be 
Distance  from  bank  0     12      3      4 

Depth  2^    5    8|    8^     10 

Distance  from  bank  10     11     12 

Depth  10    6|-      5 

Find  approximately  the  area  of  the  face  of  the  dam. 

River 

16.  From  a  straight  road  OX 
distances  are  measured,  at  inter- 
vals of  10  feet  from  0  to  80  feet, 
to  the  bank  of  a  stream  (see  Fig.) 
as  follows: 

Fig.  Ex.  16 

Dist.  along  road  0      10      20      30      40      50      60      70      80 

Dist.  tobank  0      15      22      22      20      25      22       15        0 

Find  approximately  the  area  between  the  river  and  the  ro?id. 

17.  If  the  river  of  Example  16  has  for  its  equation  200  y  -  240  x  — 
3  x2,  the  road  being  the  axis  of  x  as  shown  in  the  figure,  find  approxi- 
mately the  area  between  the  river  and  the  road,  using  the  same  in- 
tervals as  in  Example  16. 

18.  The  velocity  of  a  moving  body  at  any  time  is  given  by  the  func- 
tion V  =  4  —  P-.  Find  approximately  the  distance  traveled  in  the 
time  from  f  =  0  to  ^  =  2  at  intervals  A  ^  =  .2.  (Note,  s  =  vi\i  v  is 
constant.) 


INTEGRATION  87 

19.  The  velocity  of  a  moving  body  at  any  time  is  given  by  the  func- 
tion v  =  16  —  l^.  Find  approximately  the  distance  traveled  in  the 
time  from  ^  =  0  to  i  =  4  at  intervals  of  A  <  =  .5.  (See  note  to  Ex- 
ample 18.) 

20.  If  the  river  of  Example  16  has  for  its  equation  60  y  =  80  x  —  x\ 
as  in  Example  17,  find  the  approximate  area  between  river  and  road. 
Why  is  this  result  nearer  to  that  of  Example  16  than  the  result  of  Ex- 
ample 17?    (Suggestion:  Compare  the  three  graphs.) 


CHAPTER  VII 

INTEGRATION  AS  A  SUMMATION.  THE  DEFINITE 
INTEGRAL.  APPLICATIONS  OF  THE  DEFINITE 
INTEGRAL 

20.  Integration  as  a  Summation.    The  Definite  IntegraL 

Let  us  return  to  the  simple  function  y  =  x^  and  try  to 
generalize  the  process  discussed  in  Art.  19.  Instead  of 
passing  from  ic  =  1  to  a;  =  4  at  intervals  of  1  or  ^,  let  us 
pass  at  any  interval  A  x.  We  may  write  for  the  sum  of  the 
rectangles 

2r  =  (1)2  ^x+{l  +  ^xy^x+  {l  +  2^xy  ^x+,.,, 

+  {4.  —  ^xy  ^x 

or,  adopting  the  sigma  notation, 

X  =4 — A  a; 

Sr  =  Src^  A  x,  this  last  being  simply  a  condensed  way  of 

writing  the  sum  of  the  terms  of  the  series.  Then  as  we 
have  seen  (32)  the 

Curve-area  =  limit  Sr  =  ^^^^     Sa:^  A  x  (33) 

x=l 

If  we  can  find  the  limit  of  this  sum  we  can  find  the  curve- 
area.  If  we  generalize  again  and  begin  with  any  value  of 
X,  say  X  =  a,  and  end  with  any  other  value,  say  x  =  h, 
instead  of  oj  =  1  and  a:  =  4,  we  may  write 

Curve-area  =  jj^^o     2^'  ^  ^ 

X  =  a 

88 


INTEGRATION  89 

Thus  the  problem  before  us  is  to  find 
Hmit     ^2^2^^  =  Hmit^     \a^  A  x  +  (a  +  A  x)^  Ax 

+  {a+2AxyAx+ +  (b  —  AxyAx\  (34) 

If  we  use  dx  instead  of  A  a;  to  indicate  that  A  x  approaches 
the  limit  zero  and  use  the  old-fashioned  long  S,  /,  instead 
of  limit  S,  we  may  write  the  above 


/: 


x'dx  =  a'^dx  +  (a  +  dx^dx  -\-  (a  -{-  2  dxYdx  + 


+  (6  —  dxYdx  (35) 

We  note  that  every  term  of  the  series  is  of  the  form  x'^dx 
and  that  xHx  =  c^  (  —  )• 

Each  term  of  the  series  (34)  is  of  the  form  x^  A  x,  and 
also,  as  we  note, 

(a:  +  A  xY c^  (a) 

3  3 


(?)= 


x'^  A  X  -\'  x  A  x"  -\- 


y.2    A    ^  -L    ^  ~\~Z^      '       ^  ^ 


or 


x'^Ax  =  a(^j  —  xA^^  — 


AT^''  (b) 


Now  when  A  x  is  very  small  (approaches  0)  then  A  x^ 
is  very  small  even  as  compared  with  A  x,  and  A  x^  is  even 
smaller.  We  express  this  by  saying  that  A  x^  and  A  x^  are 
infinitesimal  even  as  compared  with  A  x,  or  that  they  are 
infinitesimals  of  higher  order.     The  equations  (b  and  a) 


above  may  then  be  written,  using  the  notation  c^  ( V )  ^^ 


© 


90  CALCULUS  AND  GRAPHS 

indicate  that  we  shall  neglect  the  terms  containing  Ax^ 
and  A  x^  as  being  so  very  small;  *  as  being,  that  is,  infin- 
itesimals of  higher  order. 


x^dx 
and 


-(f) 


x^dx  =  (^+J^^y  _^  (35) 

Thus  each  term  of  the  series  (35)  may  be  written  as  the 
difference  of  two  terms  (36)  and  the  sum  of  the  series  of 
terms  (35)  as  the  difference  of  the  sums  of  two  series  (36) 
as  follows: 

(a  +  dxydx    =v"'— ^         («  +  ^^)^ 


ia  +  2dx)^dx  = 


3 

(a 

+  2  dxy 

3 

(a 

+  3  dxy 

3 

(a  +  2  cb;y 


{h-dxYdx    =  ^  _       l^ZZ^ 

Adding  in  columns  and  noting  that  the  first  term  of  the 
positive  series  cancels  the  second  of  the  negative  series, 
the  second  of  the  positive  the  third  of  the  negative,  etc. 
we  have 

xHx  =  a^  dx-\-  {a-\-  dxY  dx  +....+  (6  —  dxY  dx 
3        3 

*  To  illustrate  this  let  the  student  put  Ax  ==  .00001  or  some  even  smaller  number 


INTEGRATION  91 

and  we  have  succeeded  in  summing  this  infinite  series  of 
infinitesimal  terms. 

Since,  however,  x^  dx  =  d  f  r-  j  we  see  that 

P'^- =/<!)  =  I 

where  the  symbol,  /,  is  the  sign  of  integration.  Therefore, 
we  may  write 

f;}'^'[P-"l-^l''i-i  (37) 

/  meaning  sum;  /  meaning  integral. 

and  since  the  symbol  /  meaning  sum  is  related  thus  to 
the  symbol  /  meaning  integration  we  call  the  expression 

/   x^  dx  Si  definite  integral  and  the  expression    /  x^  dx  an 

indefinite  integral.  The  initial  and  terminal  values  of  x 
(a  and  b)  are  called  the  lower  and  upper  limits  respectively 
and  (37)  may  be  expressed 

To  find  the  definite  integral  from  x  =  a  to  x  =  b  of  x^  dXy 
we  find  the  indefinite  integral  of  x^  dx,  in  it  substitute  first 
the  upper  limits  then  the  lower  limit  and  subtract  the  latter 
result  from  the  former  (38) 

Note:  The  student  may  say:  It  is  true  that  the  terms  neglected  in 
the  preceding  summation  are  very  small,  but  we  have  a  very  large 
nimiber  of  them.  Why  do  they  not  add  up  to  something  that  can 
not  be  neglected? 

But  he  must  remember  that  not  only  is  each  term  very  small,  but 
also  that  it  is  very  small  even  as  compared  with  the  term  just  before 
it.  Beginning  with  the  end  term,  each  term  can  be  neglected  in  com- 
parison with  the  term  just  before  it;  and,  therefore,  the  sum  of  all 
these  infinitesimal  terms  can  be  neglected  in  comparison  with  the  sum 
as  obtained  by  integration. 


92  CALCULUS  AND  GRAPHS 

To  generalize  once  more  consider  the  function  y  =  f{x). 
I'hen 

J  f(x)dx  =  f(a)dx  +  f(a  +  dx)  dx+  . , .  +  f(b  —  dx)  dx  (39) 

Let         f(x)dx  =  d  F{x)  =  F{x  +  dx)  —  F{x) 
Then  the  series  of  (39)  may  be  written 
f{a)dx  =  F(a  +  dx)  —  F{a) 
f(a  +  dx)dx  =  F(a+2  dx)  —  F(a  +  dx) 
f{a  +  2  dx)dx  =  F(a+S  dx)  —F(a+2  dx) 

f(b  —  dx)dx     =  F(b)  —  F(b  —  dx) 
Adding  we  obtain 

J  f{x)dx  =  F(b)—F{a) 

But  since  f(x)dx  =  dF(x) 

J  f(x)dx  =    F{x) 
Therefore 

/  f(x)dx  =      f(x)dx\  =  F{x)^  =  F(b)—F(a)    (40) 

and  our  theorem  (38)  is  seen  to  be  true  for  any  function, 
provided  we  can  find  its  indefinite  integral. 

Note:  It  is  frequently  desirable,  when  the  method  of  integration  by 
substitution  is  used,  to  carry  along  the  limits  of  the  new  variable  in- 
stead of  returning  to  the  original  variable.    For  example 

f«         X 
1.     f    — /  dx.    Put  x^  -\-  a^  =  u^,  X  dx  =  u  du,  and  the  integral 

•^  oV  ^2  +  a2 


INTEGRATION  93 

becomes   fdu.     But  when  x  =  0,  i^  =  a;  when  a;  =  a,  w  =  a  v  2,  so 
that  we  have 

/'«         Xdx  AV2  "|aV2 

2.     /       '^^^^  .     Put  x'^2  —  5  =  w,  a;V2  ^^o:  =  2.  du,  and  the  integral 
becomes,  since  u  =  —  5  when  x  =  0,  and  w  =  3  when  x  =  4, 


EXAMPLES 

Evaluate  the  following  definite  integrals. 


1  (5 
4  5' 


)dl 


1.     /    (x'-i +2x  — 3)(ix    2.     /     /x3  — —  j^x3.     /    a2_i6^)< 

sin  xdx  ^'    j    ^^^ 2  xdx         6.     I    sin 2  x  cos 2 xdx 

0  *7  X  «/  X 

T  T 

7.    /      -  8.    /    {y-{-y'+y')dy  9.    I   xVx'  +  ldx 

J  10?^^  J  1  Jo 

21.  General  Meaning  of  the  Definite  Integral.  For 
convenience  and  simplicity  of  presentation  we  introduced 
the  idea  of  the  definite  integral  as  the  problem  of  finding  a 
curve-area.  The  general  idea  involved  in  such  an  integral 
is,  however,  the  summation  of  an  infinite  number  of  in- 
finitesimal terms,  regardless  of  the  exact  meaning  of  these 
terms.  This  is  the  second  remarkable  intellectual  tool 
with  which  the  calculus  supplies  us.  (See  Art.  6.)  To 
make  use  of  this  new  tool  all  that  is  necessary  is  to  form 
the  element  of  the  thing  to  be  summed ;  to  get  a  differential 
expression  representing,  in  terms  of  some  independent  va- 
riable, any  one  of  the  terms  to  be  summed.    This  element 


94 


CALCULUS  AND  GRAPHS 


being  found  and  formed,  the  remainder  of  the  process  con- 
sists in  integrating,  and  substituting  the  Hmits,  as  shown 
in  (40).  In  the  following  articles  we  shall  illustrate  this 
process  by  taking  up  some  of  the  apphcations  of  integration 
as  a  summation. 

22.  Areas  of  Curves.     Let  the  graph  of  any  function, 
y  =  /(^)?  be  represented  in  Fig.  22,  and  let  it  be  required  to 


0 


M 

dx 

N 

x-a 

Fig.  22 

x=b 

find  the  area  PMNQ.  At  any  point  jB,  between  P  and  Q, 
drop  ordinates  to  the  axis  of  x  and  construct  the  rectangle 
RB  as  shown.  When  dx  is  taken  small  enough  we  may 
write  the  area  of  this  rectangle,  which  is  the  element  of 
the  thing  to  be  summed,  (the  term  the  sum  of  all  of  which 
will  give  us  the  area  required)  in  the  form 

dA  =  y  dx  =  f{x)  dx. 

Therefore,  for  the  total  area  we  have 


/b  nh 

ydx  =J  fix) 


dx 


Example.    Find  the  area  of  the  curve  y  =  1  —  x^  from 
X  =  —  1  to  a;  =  L    We  have 


dA  =  y  dx  —  (1  —  x^)  dx 


INTEGRATION  95 

and 

A  =  f\l  -  x')dx  =  x-  f^  =  (1  -  A)  -  (-  1  +  1^) 

=  1  —  ^+1  —  ^  =  1  =  1|  square  units. 

EXAMPLES 

Find  the  curve-area  for  each  of  the  following  curves  between  the 
limits  named  and  compare  the  results  with  those  of  the  examples  under 
Art.  19,  Chapter  VI. 

1.  i/  =  a:2  +  2  8.  92/  =  5+4a;  —  ^2 
a;  =  ltoaj=3  x  =  —  1,  x=5 

2.  y  =  x^  —  9  9,  y  =  2x^ 

a;  =  l,  x=2  x  =  l,  x=3 

3.  2/  =  4  —  x2  10.  2/  =  3  a;3 

x  =  — 2,  a;=2  x  =  l,  a;=3 

4.  ?/  =  9  —  x^  11.  2/  =  sin  X 

X    ^^  """  Of  X    ^  O  X   ^  -T-f  X   ^  —r" 

5.  4  2/  =  3  +  2  X  —  x2  12.  2/  =  sin  a; 
a:=0,  x=2  x  =  0,  a;='rc 


6.  4  2/  =  3  +  2  X  —  a;2  13.  2/  =  cos  a; 


aj  —  ~~~  i-y  X  —  O  X  —  ■""*  —J  37  — 


7.  9  2/  =  5  +  4  x  —  a;2  14.  2/  =  cos  x 

x=0,  a;  =  2  a;=0,  a;=x 

15.  Solve  problems  17  and  20,  Art.  19,  by  integration. 

16.  Find  the  area  between  the  two  curves  2/^  =  4  x  and  x^  =  4  y. 

17.  Find  the  area  in  the  first  quadrant  between  the  curves  y  =  sin  x, 
y  =  cos  x  and  the  axis  of  y. 

18.  Find  the  area  between  the  curves  y  =  sin  x,  y  =  cos  x  and  the 
axis  of  X  from  x  =  0  to  x  =  ^. 

19.  Find  the  area  enclosed  by  the  graphs  of  the  two  functions  y  =  x^ 
and  y  ~  X. 

V2 

20.  Find  the  area  between  the  lines  ?/  =  sin  x  and  y  = ,  from 


m 


CALCULUS  AND  GRAPHS 


21.  Find  the  area  bounded  by  the  hnes  y  ==  cosx  and  2y  =  V  2 
and  the  axis  of  x. 

22.  Find  the  area  between  the  curve  y  =  x^  and  the  line  y  =  IQ. 

23.  Find  the  area  between  the  two  lines  x'^y  =  1  and  3  a;  +  4  i/  =  7. 

23.  Volumes  of  Surfaces  of  Revolution.    Let  the  curve 
y  =  /(^)  be  revolved  about  the  axis  of  x.     (See  Fig.  23.) 


x=b       X      I 


Fig.  23 

The  small  rectangle  y  dx  will  generate  a  right  circular 
cylinder  of  radius  y  and  height  dXy  of  which  the  volume 
is  wy^  dx  =  dv.  This  is  the  element  of  volume,  dx  being 
taken  small  enough,  which  being  summed  will  give  the 
total  volume  of  the  surface  of  revolution.    Thus 


V 


4 


y'^  dx 


(41) 


If  the  curve  is  revolved  about  0  F  we  have 

'*y  =  d 


dv  =  TT  x^  dy  and 


/y  =  d 
X 
y=.c 


x'^  dy 


Example.  Find  the  volume  generated  by  revolving  the 
curve  y  =  x^  about  (a)  OX;  (b)  0  Y;  (c)  the  line  y  =  —  2; 
from  aj  =  0  to  x  =  2  (see  Fig.  19). 


INTEGRATION  97 

(a)  V  =  IT  I  y^dx  =  it  I  x^  ax  =Tr —-  \    =  — -— 

*^  0  *^  0  5  Jo         5 

(b)  y  =  TT  /  a:^^!/  =  TT  I  y  dy  =  —^      =  8  tt 

*^  0  *^  0  ^   Jo 

(c)  In  this  case  the  radius  of  the  circular  cross  section  is 

y  +  2, 

(y  +  2ydx  =  TT  /   (x2  +  2ydx 


0  *^    0 

(x^  +  4  x2  +  4)  dx 


i-2 

=    T  I     (x 


EXAMPLES 

Find  the  volume  got  by  revolving  each  of  the  following  curves  about 
the  axis  specified,  between  the  limits  given. 

1 .  2/  =  y/^  +  2    about  OX,  from  x  =  0  to  x  =  4. 

2.  y  —  X  -\-  -    about  OX,  from  a:  =  1  to  a:  =  3. 

X 

3.  3^  =  2/  +  ?/-     about  07,  from  2/  =  0  to  2/  =  4. 

4.  J  =  2/^' "  —  2     about  OY ^  from  1/  =  1  to  ?/  =  4. 

5.  y  =  y/ X  +  2     about  y  —  — -2  from  x  =  0  to  x  =  4. 

0.  ?/  =  2\/ X  +  3     about  2/  =  1  from  x  =  1  to  x  =  3. 

x2        y- 

7.  The  ciu^e  whose  equation  is  —  +  —  =  1  is  called  an  ellipse. 

Find  the  volume  got  by  revolving  the  ellipse  (1)  about  OX.    (2)  about 

or. 

8.  Given  that  the  curve  x^  +  i/^  =  a-  is  a  circle  with  radius  a,  find 
by  integration  the  volume  of  a  sphere  of  radius  a. 


98  CALCULUS  AND  GRAPHS 

9.  The  curve  xy  =  10,  called  an  hyperbola,  is  revolved  about  OY. 
Find  the  volume  generated  from  y  =  1  to  y  =  4. 

10.  Find  the  volume  generated  by  revolving  the  arch  of  the  curve 
y  —  x^  —  2x  about  the  axis  of  x. 

11.  Find  the  volume  generated  by  revolving  the  arch  of  the  curve 
X  —  2y  —  2/2  about  the  axis  of  y. 

12.  The  equation  a;^  +  ^/^  =  25  represents  a  circle  of  radius  5.  Find^. 
by  revolving  the  circle  about  OX,  the  volume  of  the  spherical  cap 
(called  segment)  from  a;  =  2  to  x  =  5. 

13.  The  circle  of  example  12  is  cut  by  the  straight  line  y  =  x.  Find 
the  volume  got  by  revolving  about  OX  the  area  bounded  by  the  circle, 
the  straight  line  and  the  axis  of  x. 

14.  Find,  by  integration,  the  volume  of  a  right  circular  cone  of 
height  h  and  angular  opening  2- 

15.  The  curve  y  =  2  v  sin  x  is  revolved  about  OX.  Find  the  volume 
generated  from  x  =  0  to  a;  =    %. 

16.  The  curve  y  =  tanx  is  revolved  about  OX.  Find  the  volume 
generated  from  a;  =  0  to  a?  =  ^. 

17.  The  curve  x  =  cos y  Vsin y  is  revolved  about  OY.  Find  the 
volume  generated  from  y  =  0  to  y  =    7C. 

18.  The  curve  y^  =  sin  x  cos  x  is  revolved  about  OX.  Find  the 
volume  generated  by  an  arch  of  the  curve. 

24.  Fluid  Pressure.  We  have  learned  in  physics  that 
the  pressure  on  a  surface  immersed  in  a  fluid  is  equal  to 
the  weight  of  the  fluid  which  rests  upon  the  surface  as  a 
base,  and  that  the  pressure  is  exerted  equally  in  all  direc- 
tions. If,  therefore,  the  surface  be  horizontal  the  pressure 
can  be  found  by  multiplying  the  area  of  the  surface  by  the 
depth  of  the  fluid,  and  multiplying  this  product  by  the 
weight  of  the  fluid  per  cubic  unit.  But  if  the  surface  the 
pressure  on  which  we  wish  to  measure  is  not  horizontal, 
the  depth  of  the  fluid  resting  upon  it  is  not  constant.  This 
is,  therefore,  a  case  in  which  we  have  to  form  the  element 


INTEGRATION 


99 


of  the  thing  to  be  summed,  and  to  perform  the  summation 
by  integration.  We  accomplish  this  by  taking  a  strip  of 
the  area  of  the  immersed  surface,  so  narrow  that  the  whole 
strip  can  be  regarded  as  at  the  same  depth  in  the  fluid. 
Thus,  in  Fig.  24,  let  0  F  be  the  surface  of  the  fluid,  whose 


Fio.  24 


density  is  co,  and  let  the  positive  axis  of  X  be  taken  down- 
ward.   Take  any  horizontal  strip  of  area  AB  of  width  dx. 

Let  MB  =  2/2  and  MA  =  yi.    Then  we  have 
Element  of  area  =  dA  =  (2/2  —  yi)  dx 
Element  of  volume  =  dV  =  x(y2  —  yi)  dx 
Element  of  pressure  ==  dp  =  cox(t/2  —  Vi)  dx 

and  the  pressure 


0)  xivi  —  2/1)  dx 


(42) 


the  values  x  =  a  and  x  =  6  represent  the  highest  and 
lowest  points,  respectively,  of  the  immersed  surface. 

The  density,  co,  is  usually  a  constant  (for  water  co  =  62.5 
lbs.  per  cubic  foot)  but  it  may  be  a  variable. 


100                   CALCULUS  AND  GRAPHS 
O  10^ B 


A 


X 


M 


N 


Y  Example  1,  What  is  the 
pressure  on  a  rectangular 
flood-gate  6  feet  deep  and 
10  feet  broad  if  the  sur- 
face of  the  water  is  level 
with  the  top  of  the  gate? 
See  Fig. 


Fig.  Ex.  1 

In  this  case     dA  =  MN  dx 
=  10  dx 
dv  =  X  .  10  dx 
dp  =  (ji)x  10  dx 


and  p 


--  I  lOco 


X  dx  =  10  CO 


2  Jo 


11250  lbs. 


5|  tons. 


Example  2.  Suppose  the  gate  of  Example  1  is  the  end  of 
a  closed  sluice,  and  that  the  level  of  the  water  is  50  feet 
above  the  top  of  the  gate;  then 

dA  =  10  dx,   dv  =  (50  +  x)  10  dx,   dp  =  co  (50  +  x)  10  dx 
and 

p  =  I  10  CO  (50  +  a;)  do:  =  10  CO    50  a;  +  ^ 

^  0  [  ^   )  0 


=  99|  tons. 


Example  3.  Suppose  the  water,  in  Example  1,  rises  to 
within  2  feet  of  the  top  of  the  gate;  then,  as  before,  dp  = 
lOo)  X  dx  and 


=J  10  coo; 


dx  =  10  0) 


(f)> 


5  tons. 


Example  4-  Suppose  the  water  in  Example  1  is  so  mixed 
with  oil  discharged  from  a  factory  that  the  density  varies 


INTEGRATION  (/  101 

as  the  depth  below  the  surface,  being  62.5  lbs.  per  cubic 
foot  at  a  depth  of  6  feet.    Then 

62  5 
0)  =  Kx  and  co  =  62.5  when  a:  =  6,  so  that  K  =  —^  and 

o 

CO  =  —^  X,    We  thus  have 


6 


dp  =  —^  X  .  X  .  10  dx 


and 

p  =  ^  Hx'dx  =  ^flT^  7500  lbs.  =  3|  tons, 
o  •^  0  b  \6  /o 

EXAMPLES 

Find  the  pressure  on  an  equilateral  triangle,  of  10  foot  side,  immersed 
vertically  in  water,  under  the  conditions  named  in  Examples  1-4. 

1 .  When  the  base  is  in  the  surface  of  the  water. 

2.  When  the  vertex  is  in  the  surface  of  the  water  and  the  base  hori- 
zontal. 

3.  When  the  base  is  horizontal  and  2  feet  below  the  surface  of  the 
water,  the  vertex  being  downwards. 

4.  When  the  triangle  is  in  the  position  of  Example  2  except  that 
the  vertex  is  4  feet  below  the  surface  of  the  water. 

5.  A  triangle  with  base  4  feet  and  altitude  5  feet  is  immersed  verti- 
cally with  its  base  in  the  surface  of  the  water.  Find  the  pressure  on 
the  triangle. 

6.  Find  the  pressure  on  the  triangle  of  Example  5  when  the  triangle 
is  inverted. 

7.  A  trough  has  vertical  ends  in  the  form  of  a  right  triangle  with 
the  hypothenuse  horizontal,  each  leg  of  the  triangle  being  14  inches. 
The  trough  contains  mercury  to  within  an  inch  of  the  top.  Find  the 
pressure  on  one  end  of  the  trough.  (Take  the  specific  gravity  of  mer- 
cury as  13;  i.  e.,  1  cubic  foot  of  mercury  weighs  62.5  X  13  pounds.) 

8.  A  sluice  is  closed  by  a  water-gate  in  the  fonn  of  a  trapezoid,  upper 
base  6  feet,  lower  l)ase  4  feet,  and  with  sides  equally  inclined  to  the 


102  CALCULUS  AND  GRAPHS 

base.  Find  the  pressure  on  the  gate  when  the  sluice  is  full,  if  the  level 
of  the  pond  which  fills  the  sluice  is  40  feet  above  the  top  of  the  gate, 
and  the  gate  is  3  feet  high. 

9.  A  right  triangle  with  sides  3,  4  and  5  feet  respectively  is  immersed 
vertically  with  the  3  foot  side  in  the  surface  of  the  water.  What  is 
the  pressure  on  the  triangle?  To  what  depth  must  the  triangle  be 
sunk,  keeping  the  3  foot  side  parallel  to  the  surface  of  the  water,  to 
double  the  pressure? 

10.  To  what  height  must  the  triangle  in  Example  9  be  raised  to 
halve  the  pressure? 

11.  A  square  of  side  a  feet  is  placed  in  water  with  its  diagonal  vertical, 
and  a  corner  in  the  surface.    Find  the  pressure  on  the  square. 

12.  Taking  the  surface  of  the  water  as  the  axis  of  y  and  the  vertical 
line  downward  as  the  axis  of  x,  find  the  pressure  on  the  triangle  formed 
by  the  axis  of  y  and  the  two  lines  2x  -\-  4:y  =  8  and  3a;  —  4 2/  =  12. 

13.  Using  the  same  axes  as  in  Example  12  sketch  the  graph  of  y^  = 
8  X,  and  find  the  pressure  on  the  area  bounded  by  the  curve  and  the 
line  X  =  8. 

14.  Taking  the  axis  of  y  horizontal  and  the  axis  of  x  vertically  down- 
ward sketch  the  graph  of  the  curve  y^  -\-  IQ  x  =  0.  If  the  surface  of 
the  water  is  in  the  line  x  =  —  3,  find  the  pressure  on  the  area  bounded 
by  the  curve. 

15.  A  tank  is  rectangular  in  shape,  10  feet  long,  6  feet  wide  and  5 
feet  deep.  It  is  filled  with  mixed  oils  such  that  the  density  is  propor- 
tional to  the  depth  below  the  surface,  the  density  being  15  pounds 
when  halfway  down.  Find  the  pressure  on  the  ends  and  sides  of  the 
tank. 

16.  A  rectangular  dam  100  feet  wide  and  30  feet  deep  has  its  top 
1000  feet  below  the  surface  of  the  water  in  the  reservoir  which  supplies 
the  water.    Find  the  pressure  on  the  dam  in  tons. 

25.  Moment  of  Force  or  Turning  Moment.  Moment  of 
force,  or  turning  moment,  is  the  product  of  the  force  by 
the  arm;  the  arm  being  the  distance  from  the  axis  about 
which  the  force  acts  to  the  point  of  application  of  the  force. 

Let  us  suppose  that  the  force  with  which  we  are  dealing 
is  fluid  pressure  and  that  the  axis  about  which  it  acts  is 


INTEGRATION  103 

the  surface  of  the  fluid,  then  (see  Fig.  24)  the  force  acting 
on  any  Httle  strip  of  area  AB  is  the  pressure  on  that  strip. 
We  have  found  this  to  be 

dp  =  cox  {y2  —  2/1)  dx. 

The  moment  of  this  force  is,  therefore,  dM  =  x  dp  = 
CO  x^  (2/2  —  ^1)  dxy  since  the  arm  is  x,  the  distance  from  the 
surface.  Then  we  have  for  the  moment  of  force  about  the 
surface 


X  dp  =   I    CO  x^  (1/2  —  2/1)  d^ 


M  =    I    xdp  =    I    CO  x2  (1/2  —  2/1)  dx.  (43) 

Example  1.  Find  the  moment  of  force  about  the  surface 
of  the  water  of  the  rectangle  in  Example  1,  Art.  24. 

Here  dp  =  10  cox  dx 

and  dM  =  10  co  x^dx 

M  =  10  col    x'^dx  =  10  co^\=  45,000  ft.  lbs. 
^  0  o  Jo 

Example  2,  Find  the  moment  of  force  in  Example  1  about 
the  bottom  of  the  rectangle. 

In  this  case  the  arm  is  6  —  x  and 


dM  =  10  CO  x  (6  —  x)  dx 

t 
3 


M  =  10coJ  (6x  — x2)da:  =  10co|3a:2  — 


=22,500ft.lbs. 

0  \ 


EXAMPLES 


1.  Find  the  moment  of  force  about  the  surface  in  Examples  1-6  of 
Art.  24. 

2.  Find  the  moment  of  force  about  the  top  of  the  gate  in  Example  8 
of  Art.  24. 


104  CALCULUS  AND  GRAPHS 

3.  Find  the  moment  of  force  about  the  bottom  of  the  gate  in  Ex- 
ample 8  of  Art.  24. 

4.  Using  the  triangle  of  Example  9,  Art.  24,  find  the  moment  of 
force  about  the  surface. 

26.  Distance  Determined  from  Velocity,  Acceleration, 
and  Time.    We  have  seen  (Arts.  11  and  12)  that  the  velocity 

of  a  moving  body  is  given  by  «;  =  —- ,  whence  ds  =  v  dt. 

It  follows,  therefore,  that  the  distance  is  given  by 

S=Jvdt  (44) 

Also  since  angular  velocity  is  given  by 

0)  =  —r-  or  du  =  0)  dt 
at 

we  have 

e=Jo)dt  (45) 

Since,  however,  in  a  rotating  body  we  know  that  S  =  a  6, 

then  ~  =  a  — r-.    We  thus  have,  for  the  distance  traveled 
dt  dt 

by  a  point  on  a  revolving  circle  of  radius  a, 

S=ladd=iao:dt  =  aio^dt  (46) 

Example  1,  A  body  moves  with  a  velocity  v  =  3  —  2t, 
Find  the  distance  traveled  during  the  first  3  seconds;  during 
the  5th  second. 

Here 


V  = 

dt  ~'^~ 

-2t,  or  ds 

=  (3 

—  2t)dt 

Therefore, 

Sl 

-sy 

-2t)dt  = 

3t  — 

-3 

■f       =   0 
_0 

and 

S2 

-ly 

-20  dt  = 

3t- 

-15 

_4 

INTEGRATION 


105 


These  results  seem  peculiar.    Let  us  analyze  them.    We 

ds 
find  §2  to  be  negative.    But  during  the  5th  second  v  =  — 

at 

=  3  —  2t  is  negative,  which  means  that  s  decreases  as  t 

increases,  or  that  the  body  is  moving  backwards;  that  is, 

in  the  negative  direction.    The  body  actually  moves  through 

a  distance  6,  however,  during  the  5th  second. 

It  is  found  that  si  is  zero.    This  does  not  mean  that  the 

body  has  not  moved  through  any  distance  in  the  first  3 

seconds,  but  that  the  algebraic  sum  of  those  distances,  some 

positive  and  some  negative,  is  zero.    Putting 


ds 


2  ^  =  0  we  find  ^  =  I 


Then 


t 

1 

.1 

2 

2 

ds 

dt 

+ 

0 

— 

s 

incr. 

deer. 

That  is,  for  the  first  1^  seconds  s  is  increasing  and  the  body 
moves  forward.  From  1^  seconds  to  3  seconds  s  is  de- 
creasing and  the  body  moves  backward.  Therefore,  tak- 
ing the  two  periods  separately,  we  have 

(3  —  2t)dt  =  3t  —  t''\    =1 
0  Jo 

s"  =  I     (3  —  2t)dt  =  St  —  t'n    =  —  I 


s"  =  I  =  4|,  distance  traveled  during  first 


and  sc=  s' 
3  seconds. 

If  we  plot  the  graph  of  i;  =  3  —  2t  (Fig.  25)  beginning 


106 


CALCULUS  AND  GRAPHS 


with  ^  =  0,  we  see  that  the  area  between  OF,  the  line 
ABD,  and  OT  represents  the  distance  traveled  in  any  given 

V  dt  where   j   v  dt  means 
h  *^  ti 

the  limit  of  the  sum  of  the  rectangles  v  At    From  ^  =  0  to 
^  =  3  the  area  is 

AOB  =  -1 05  X  04  =  I,  plus  BCD  =  ^BCXCD  =  —^ 
The  numerical  value  of  the  total  area  is  |  +  |  =  4|. 


FiQ.  25 


Also,  the  trapezoid 
MNQP  =  i  (MP  +  NQ)  X  MN  =  ^  (—  5  —  7)  X  1  = 
—  6  is  the  distance  traveled  from  ^  =  4  to  ^  =  5.  This 
gives,  graphically,  the  results  already  obtained.  We  may 
note  also  that  when  the  velocity-time  area  is  above  OT 
the  distance  is  positive;  when  below  OTy  the  distance  is 
negative;  and  that  the  moving  particle  comes  to  rest 
{v  =  0)  at  the  point  B  {t  =  1^)  where  the  graph  crosses 


INTEGRATION  107 

the  axis  of  T.    Similarly,  if  /  =  F  (0,  where  /  is  accelera- 
tion, be  plotted,  areas  as  above  represent  velocities,  since 

-n  —  f  ^^^  dv  =  f  dt. 

Example  2.  A  wheel  revolves  until  friction  stops  it,  with 
angular  velocity  co  =  15  —  3  <.  Through  what  angle  will 
the  wheel  turn  in  2  seconds?   In  5  seconds?    In  6  seconds? 

Here 


and 


CO  =  4r-  =  15  —  3  i;  6i  0  =  (15  —  3  0  d« 
at 

ei=l{l5  —  3t)dt  =  15t  —  ^1  =  24  radians. 
^0  ^  Jo 

(15  —  S  t)  dt  =  15  t  —  ^  \  =  37^  radians. 

0  ii  Jo 

{15  —  St)dt=15t  —  ^\  =36  radians. 
0  ^  Jo 

The  value  of  ds  is  absurd  because  as  a  matter  of  fact  the 
wheel  stops  in  5  seconds.  If  the  law  co  =  15  —  3t  held 
after  5  seconds  the  motion  would  be  reversed,  and  from 
<  =  5  to  i  =  6  we  should  find  d  =  —  1^.  It  should  be 
noted  that  these  results  could  be  got  graphically  as  in 
Example  1. 

Example  3.  A  rifle  ball  is  fired  through  a  3-inch  plank, 
the  resistance  of  which  causes  an  unknown  constant  re- 
tardation of  its  velocity.  Its  velocity  on  entering  the 
plank  is  1000  feet  a  second,  and  on  leaving  the  plank  is 
500  feet  a  second.  How  long  does  it  take  the  ball  to 
traverse  the  plank?  How  thick  must  the  plank  be  to  stop 
the  ball? 


108  CALCULUS  AND  GRAPHS 

In  this  case  the  retardation  is  a  negative  acceleration  so 
that 

^  "  df  dt"" 
and 

V  =  —  Kt+  c 

But  V  =  1000  when  t  =  0,  therefore,  c  =  1000  and 

V  =  ^^  =  —  Kt+  1000. 
dt 

Also  V  =  500  when  t  =  T  (the  time  of  leaving  the  plank), 
therefore, 

500  =  —KT  +  1000  or  K  =  ^  ^^^ 

Again,  since 

^  =  —  Kt+  1000 
at 

we  have 

s  =  J  (—  Kt  +  1000)  dt  =  —  ^  +  1000  t  +  c' 

But  s  =  0  when  ^  =  0;  therefore  c'  =  0. 
Also  s  =  3  inches  =  ^  foot  when  /  =  T, 

Therefore, 

1  KT^ 

7  =  —  ^    +  1000  T 

4  2 

In  this  substitute  K  =  ^'— -  and  we  have 

1  =  _?|2.  r^+iooor 

4  T 

T  =  ^  second. 


INTEGRATION  109 , 

To  uiiswcr  tlic  second  question  we  put  v  =  0  when  t  =  T' 
in  the  general  expression  for  v.    Thus 


0  =  —  KT'  +  1000;  T'  = 

„   .    .         „       500       500      ,  K^^nAA 
But  since  K  =  -=-  =  — —  =  1,500,000 

3000 
Therefore, 


1000 
K 


rr,  1000      1      . 

^  =  i;5oo;oro  =  isoo  ^"^°'^^- 


A>so 


s  =  _  AH'  +  1000  r  =     1-500,000       1 


2   '  2    ■  (1500)2 

EXAMPLES 

1 .  If  ?;  =  5  +  7  /,  where  v  is  the  velocity  in  feet  per  second  and  t 
is  the  time  in  seconds,  find  the  distance  traveled  (a)  in  the  first  5  sec- 
onds, (b)  during  the  lOth  second. 

2.  If  V  =  3  l^y  find  the;  distance  traveled  during  the  5th  second. 

3.  If  V  =  2  t^  -\-  S  t  -{•  5,  find  the  distance  traveled  between  the 
ends  of  the  3d  and  10th  seconds. 

4.  The  speed  of  a  body  at  the  end  of  t  seconds  from  a  fixed  instant 
is  given  by  v  =  u  -{-  at  (where  u  and  a  are  constants).  Show  that  the 
distance  traveled  in  these  t  seconds  is  w/  +  ^  at^-  Show  also  that  u 
is  the  speed  at  the  fixed  instant,  and  that  the  acceleration  is  constant 
and  equal  to  a. 

5.  Find  the  number  of  revolutions  made  in  the  first  5  minutes  by 

a  wheel  which  moves  with  an  angular  speed  CO  =  radians  per 

1000  ^ 

second. 

6.  In  Example  5,  find  the  number  of  revolutions  made  during  the 
3d  minute. 


110  CALCULUS  AND  GRAPHS 

7.  The  velocity  of  a  moving  body  is  given  by  the  equation  v  = 
2  cos  t.  How  long  and  how  far  will  the  body  move  before  it  comes  to 
rest? 

8.  How  far  will  the  body  of  Example  7  move  in  iu  seconds? 

9.  A  boy  starts  a  slide  on  ice  with  a  speed  of  14§  feet  per  second. 
He  stops  in  3f  seconds  after  sliding  26f  feet.  His  speed  at  any  time 
is  given  by  the  equation  v  =  —  32  Kt  -\-  14|-  where  K  is  the  coefficient 
of  friction  between  the  boy's  shoes  and  the  ice.    Find  the  value  of  K. 

10.  If  the  boy  of  Example  9  starts  with  a  speed  of  20  feet  per  second, 
how  long  and  how  far  will  he  slide  before  stopping? 

In  each  of  the  following  examples  find  how  long  and  how  far  the 
body  will  move  before  it  comes  to  rest  (s  in  feet;  t  in  seconds).  Check 
your  results  by  means  of  a  graph. 

ll.v  =  5  —  4:L  12.  y=  4^  —  5.  13.^=^^  —  2. 

14.  /  =  2  —  t;  starting  from  rest  (/  =  acceleration). 

15.  /  =  3  —  2  t;  starting  from  rest. 

16.  /  =  ^t  —  4;  starting  from  rest. 

How  long  and  through  what  angle  will  a  wheel  turn  before  coming 
to  rest  if  its  motion  is  given  by  the  following  equations?  Check  your 
results  by  means  of  a  graph.  If  the  radius  of  the  wheel  is  2  feet,  how 
far  will  a  point  on  the  rim  of  the  wheel  travel? 

17.  CO  =  3  —  4  <.  18.  CO  =  ^  —  «.  19.  CO  =  2  i  —  3. 

20.  a  =  1  —  4  t;  starting  from  rest  ( a  =  angular  acceleration) 

21.  a  =  X  —  4:t;  starting  from  rest. 

22.  a  =  t  —  f ;  starting  from  rest. 

23.  A  torpedo,  shot  under  water,  has  a  speed  of  50  feet  per  second 
at  the  moment  its  compressed  air  is  exhausted.  It  suffers  a  constant 
retardation  of  5  feet  per  second  per  second.  How  long  and  how  far  will 
it  travel  before  coming  to  rest? 

24.  A  torpedo  shot  under  water  has  a  speed  of  60  feet  per  second 
at  the  moment  its  compressed  air  is  exhausted.  It  comes  to  rest  after 
traveling  350  feet.  How  long  does  it  travel  before  coming  to  rest? 
What  is  the  retardation? 

25.  A  rifle  bullet  enters  a  sand  bank  300  feet  thick  with  a  velocity  of 
1000  feet  per  second.  If  it  suffers  a  retardation  of  2000  feet  per  second 
per  second,  how  far  will  it  penetrate  the  bank  before  coming  to  rest? 


INTEGRATION  111 

26.  How  long  will  the  bullet  of  Example  25  take  to  travel  its  whole 
path  through  the  bank?  The  first  half  of  the  path?  The  second  half? 
With  what  velocity  does  the  bullet  pass  the  midpoint  of  its  path? 

We  shall  conclude  this  portion  of  our  work  with  the  dis- 
cussion of  a  few  more  applications  of  integration  as  a  sum- 
mation; cases  in  which  we  sum  an  infinite  number  of  in- 
finitesimal terms,  or  to  express  the  idea  a  little  differently, 
in  which  we  find  the  limit  of  the  sum  of  a  series  of  very 
small  terms,  each  of  a  certain  type,  as  the  number  of  terms 
becomes  very  great.  Our  first  task  in  each  case,  as  will  be 
seen,  will  be,  as  in  the  preceding  applications,  to  form  the 
type  term  of  the  series;  the  element  of  the  thing  to  be 
summed. 

27.  Center  of  Pressure  and  Center  of  Gravity.  Return- 
ing to  the  discussion  of  Art.  25,  let  us  suppose  that  there 
is  a  certain  distance,  J,  below  the  axis  of  y  such  that,  if 
the  total  pressure  were  appHed  at  that  distance,  the  mo- 
ment of  force  would  be  the  same  as  that  already  computed. 
Then 


and 


o)x  {y2  —  Vi)  dx  =  I    o)x^  (2/2  —  Vi)  dx 

/X2 
CO  ^2  (?/2  —  yi)  dx 


X  /•X2 


"/: 


(47) 


CO  X  (^2  —  2/1)  dx 


Such  a  distance  is  called  the  Center  of  Pressure  of  the  sur- 
face. It  is  usually  measured  from  the  surface  level  of  the 
fluid  causing  the  pressure. 

Again,  if  the  force  of  which  the  moment  is  to  be  found, 
be  gravity,  a  point  can  be  found  such  that  if  the  total  force 
be  applied  at  that  point  the  turning  moment  will  be  equal 
to  the  moment  of  force  as  found  by  the  methods  of  Art.  25. 


112 


CALCULUS  AND  GRAPHS 


Such  a  point  is  called  the  center  of  gravity  of  the  body. 
Thus  given  the  closed  area  bounded  by  F{x,  y)  =  0,  where 
the  equation  may  be  solved  for  ?/  as  a  function  oi  x\y  = 
fi(x)y  say)  or  for  a:  as  a  function  oi  y  \x  =  f2{y)i  say), 
we  may  take  an  element  of  area  parallel  to  0  F  or  one 
parallel  to  OZ  as  in  Fig.  26.     We  thus  have  in  the  two 


Y 

1 
1 

1 
1 
1 

t 

1 
1 

1 

X 

1         ! 

1 

1              1     X 

0 

a 

d 

X             b 

FiQ.  26 


cases  respectively,  m  being  the  mass,  p  the  density,  or  mass 
per  square  unit,  and  g  the  acceleration  due  to  gravity. 


dA  =  (2/2  —  2/1)  dx 

dm  =  p  (2/2  —  yd  dx 

d  Force  =  g  p  (y2  —  yd  dx 

d  Moment  =  g px  {y^  —  2/1)  dx 


dA  =  {x2  —  x-j)  dy 
dm  =  p  {x2  —  ^1)  dy 
dF  =  gp  (0:2  — :ri)  dy 
dM  =  gpy  (x2  —  Xi)  dy 


where,  in  the  former,  the  force  of  gravity  acts  downward 
parallel  to  0  F,  and  in  the  latter  to  the  right  parallel  to  0  X. 
Therefore 


Mo.  of  Force  = 


i: 


=b  Py=d 

gpx(y2-yi)dx,or  j       gpy(x2—Xi)dy 


INTEGRATION 


113 


But  if  the  center  of  gravity  be  the  point  (X,  Y)  then,  the 
total  force  being 


F 

we  have 


and 


dF  =  J        gp(ij2  — Vi)  dx,  or  J        gp{x2  —  x)dy, 

__  nb  f*b 

Xf  gp(y2  —  yi)dx  =  l  gpx(y2  —  yi)dx 

^    a  *^    a 

/d  nd 

gp{^2  —  xi)  dy  =  J   gpy  {x^  —  Xi)  dy 


or 

^6  nd 

gpx(y2  —  yi)dx  J  gpy(x2  —  xi)dy 

X=  -7^ and  7  =  -jh (48) 


,/: 


gp  (2/2  —  2/1)  dx 


f: 


gp(x2  —  xi)dy 


From  which  the  common  factor  g  may  be  removed,  and 
also  p  if  the  density  be  constant. 


Example  1.  Find  the  cen- 
ter of  pressure  of  an  equila- 
teral triangle  of  side  4  feet, 
immersed  vertically  in  water 
with  a  side  in  the  surface  of 
the  water.    See  Fig. 


Here 


0J5  =  2\/8 


and 


114 


CALCULUS  AND  GRAPHS 


AC 
4 

=  — ^^ — -TT-  or  4C  =  —p  (2  V3  —  a;)  =4 rr 

lerefo 

'          d  Area  =  (  4  —    /i)  ^^ 

dp      =cox(4-:^) 


c?a; 


dMo.    =  o)x^ 


dx 


and 


__  AVs        /         2x\  AV3         /         2x\ 


Whence 


/W3     /         2a;\  AV3/  2  o^^X 


/rn 


2a:\ 


/r\ 


2x^\ 


4a;s 


2  V3, 


2a;2 


2x^- 
3  Vs. 


2V3 
0 


2V3 
0 


V3 


Example  2.  A  triangle  is 
formed  by  the  Unes  y  =  x  \/3, 
2/  =  4  \/3  —  X  a/3  and  the 
axis  of  X.  Find  the  center 
of  gravity  of  the  triangle. 
See  Fig. 

Here  dA  =  y  dx  for  each  of 
the  two  right  triangles  formed 
as  shown  in  the  figure. 


INTEGRATION 

dA  =  X  \/3  dx  or  dA  =  (4  \/s  —  x  \/3)  dx 


115 


dm  =  p  x  \/3  dx  or  p  (4  \/3  —  x  \/3)  cte 
dF  =  g  p  X  \/3  dx  OT  g  p  (4  \/3  —  x  \/3)  c?a: 
dAf  0  =  g  p  x^  \/3  da:  or  g  px  {4:  \/3  —  x  \/3)  ^^ 
and  the  total  moment  of  force  is 

gpx^\^3dx+l    gpx  {4:\/s  —  x\/3)  dx 
0  •^  2 

The  total  force  is 

F  =  /  gpx  \/3  6/x  +  /   fif P  (4  \/3  —  X  \/3)  cia; 

*^    0  '^2 

Therefore,  assuming  the  density,  p,  to  be  constant, 

Z    {7pJ   x^/sdx+gpj    (4:\/S  —  x\^)dxl 

=  gp  f  x^Vsdx+gpf  x{4:^/3  —  x  \/3) 

/x^\/Sdx+   I  x(4:\/S  —  x  Vs)  dx 

/x  VS  cia;  +   /  (4  \/3  —  x  V3)  dx 
0  *^  2 


dx 


or 


X 


vvr 


2      r 


Jo 


■Vs' 


+ 


+ 


2\/3a;2 


vri* 


J^    8\/3  ^ 


4v-3.-^-^^T^  ^^^ 


L     2     Jo      ^  -      ^. 

as  could  have  been  seen  from  the  symmetry  of  the  figure. 
To  find  y  we  have 

dA  =  ix,-x{)  #  =  [(4  -  :^)  -^]d2/  =  (4  -  ^)# 


116  CALCULUS  AND  GRAPHS 

dm  =  pU  —  ^j  dy 

dF  =  9p\A  —  :^dy 

/  2y\ 

dMo  =  sf  p  2/  r  4  —  7/|  J  dy,  and 

j2V3    /  2y\ 

r2V3/  2  2/\ 


f''U^H\...     ~4V3 


EXAMPLES 

1.  Find  the  center  of  pressure  in  Examples  1-6,  8  and  9  of  Art.  24. 

2.  Find  the  center  of  gravity  of  a  rectangle  8  inches  long  and  3  inches 
high. 

3.  Find  the  center  of  gravity  of  a  right  triangle  with  sides  3,  4  and 
5  feet,  the  4-foot  side  being  along  OX. 

4.  Find  the  center  of  gravity  of  the  triangle  of  Example  3  if  the  3-  j 
foot  side  is  along  OX.  j 

5.  Find  the  center  of  gravity  of  the  triangle  of  Example  3  if  the  j 
hypothenuse  is  along  OX.  | 

6.  Find  the  center  of  gravity  of  the  area  bounded  by  the  curve 
y  =  ^x  and  the  straight  line  x  =  4. 

7.  Find  the  center  of  gravity  of  the  area  bounded  by  the  curve  < 
y  —  x^  and  the  line  2/  =  4. 

8.  Find  the  center  of  gravity  of  the  area  bounded  by  the  curve  I 
y  =  x^j  the  line  x  =  2  and  the  axis  oi  x.  | 

9.  An  equilateral  triangle,  each  side  of  length  2  a,  has  one  side  on  \ 
the  axis  of  x,  the  origin  being  at  the  middle  Doint  of  that  side.  Find  ; 
the  center  of  gravity  of  the  triangle. 

10.  Find  the  center  of  gravity  of  the  area  between  the  two  curves 
y  =  x"^  and  y"^  =  x. 


INTEGRATION  117 

11.  Find  the  center  of  gravity  of  the  area  between  the  curve  y  ^  x^ 
and  the  line  y  =  x. 

12.  Find  the  center  of  gravity  of  the  area  between  the  curve  y^  =  x 
and  the  Hne  y  =  x. 

13.  Find  the  center  of  gravity  of  the  area  between  the  lines  y  =  x 
and  y  =  2x  from  the  origin  to  the  line  x  =  4. 

28.  Mean  Value.  The  average  value  of  a  set  of  quantities 
is  the  sum  of  the  quantities  divided  by  their  number.  For 
example,  the  average  value  of  the  record  of  a  student  whose 
tests  were  marked  25,  60,  60,  75,  100,  80,  40  per  cent., 
would  be 

25+60+60+75+100+80+40      440       ^^. 

=  -y-  =  62y  per  cent. 

The  mean  value  of  a  set  of  quantities  is  the  limit  of  the 
average  value  as  the  number  of  quantities  is  indefinitely 
increased.  Thus,  if /(x)  represent  any  one  of  the  quantities 
and  n  the  number  of  quantities,  then 

Average  value  of /(x)  =  — =^-^  ^ 

71 

Mean  value  of /(a:)  =  limit  _Ml)  (^0) 

n=QO         7j 

The  form  of  (50)  suggests  an  integral,  but  we  can  not  pass 
to  the  limit  and  integrate  without  knowing  with  respect 
to  what  variable  we  integrate.  The  average  value  of  f{x) 
(49)  can  obviously  be  written: 

^  fix)  Ax       ^  fix)  Ay 

A '  ~X '     etc. 

n  A  X  n  A  y 

and  the  mean  value 

limit  S/W^^     limit  M^iAiA  ,tc 
»*  =  °°      n  Ax  '» ^  °°      n  Ay 


118 
or 


CALCULUS  AND  GRAPHS 


/'X2  f*  2/2 

f{x)dx       /    }{x)dy 

/►X2 
J"" 


Ay 


■'  etc. 


since 


,  obviously  li™i*^  nLx  =  J"^^^  ^^^ 


(51) 


In  the  case  when  we  multiply  numerator  and  denominator  j 
hy  Ay  and  obtain  the  second  form  of  (51)  the  relationJ 
between  the  variables  x  and  y  must  be  known  or  capable! 
of  determination.  i 


Example.  What  is  the  mean  value  of  the  ordinate  of  the  \ 
curve  y  =  x^j  from  the  point  (0,  0)  to  the  point  (2,  4)?         ; 


M.  V.  =  1™!^ 


Sa:2 


Shall  we  multiply  numerator  and  denominator  hy  A  x,  A  y  \ 
or  by  some  other  increment  before  integrating?  The  choice  i 
of  the  increment  depends  upon  how  we  select  the  ordinates.  j 
If  we  space  the  ordinates  equally  along  OX,  we  should] 
multiply  by  A  x;  if  equally  along  0  F,  by  A  y,  etc.  In  the  \ 
former  case  we  should  have  I 


M.  V.  = 


/, 


2 

x'^  dx 


I  dx 

^  0 


In  the  latter  case 


M.  V.  =  -^^ 


/2/  =  4  /*4 

x'^dy        f  y 


JP    _     4     _     11 

2    ~   a   ~   ^^ 
0 


y' 


0       =  -Ajp  _  2 


dy 


f'dy 
^  0 


y 


INTEGRATION 


119 


We  see  that  the  mean  value  of  the  ordinate  (or  of  any  func- 
tion) is  dependent  upon  the  manner  in  which  values  of  the 
ordinate  (or  function)  are  chosen.  Fig.  27  illustrates  this 
in  the  case  of  the  example  just  worked.  It  will  be  noted 
that  spacing  the  points  equally  along  0  Y  crowds  the  or- 
dinates  together  where  they  are  longest. 


Y 

*  4- 

i 

3- 

/ 

2 

/ 

/ 

1 

/ 

X 

0 

r 

> 

F 

4- 

3- 

\ 

i 

1 

o. 

1 

1- 

( 

X 

0 

z 

Fig.  27 


EXAMPLES 

1.  Find  the  mean  value  of  the  ordinate  of  the  curve  y  =  x^^  from 
the  point  (1,  1)  to  the  point  (2,  8),  (a)  equally  spaced  along  OX;  (b) 
equally  spaced  along  OY , 

2.  Find  the  mean  value  of  the  ordinate  of  an  arch  of  the  curve 
y  =  sin  X,  spaced  with  respect  to  x. 

3.  A  particle  moves  in  a  straight  line  according  to  the  law  s  = 
5  -{-  Qt  —  3t^,  Find  the  mean  value  of  the  velocity  during  the  first 
second. 

4.  Find  the  mean  value  of  the  velocity  of  the  particle  in  Example  3 
between  the  distances  5  feet  and  8  feet. 

5.  Find  the  mean  distance  from  the  origin  at  intervals  of  time  during 
the  first  second,  of  the  particle  in  Example  3. 


120  CALCULUS  AND  GRAPHS 

6.  A  particle  moves  in  a  straight  line  according  to  the  law  s  = 
41  —  t^.  Find  the  mean  distance  from  the  origin  during  the  first  two 
seconds;  during  the  fifth  second. 

7.  Find  the  mean  value  of  the  velocity  of  the  particle  in  Example  6 
during  the  first  two  seconds;  during  the  fifth  second. 

8.  A  particle  moves  in  a  straight  line  with  a  velocity  v  =  32  ^  +  3, 
starting  at  the  origin.  Find  the  mean  velocity  of  the  particle  during 
the  first  three  seconds.  If  the  particle  had  moved  with  a  constant 
velocity  equal  to  the  mean,  how  far  would  it  have  moved  in  the  three 
seconds?    How  far  did  it  actually  move? 

9.  Find  the  mean  value  of  the  velocity  with  respect  to  the  distance 
of  the  particle  in  Example  8. 

10.  Oil  is  poured  into  a  tank  containing  water  so  that  the  density 
(mass  per  cubic  unit)  of  the  liquid  is  decreasing  at  the  rate  of  0.2  lbs. 
per  minute.  Find  the  density  of  the  mixture  at  the  end  of  5  minutes. 
Find  the  mean  value  of  the  density  during  the  5  minutes. 

11.  Find  the  mean  ordinate  of  y  =  x^  —  4  a;  +  3  between  x  =  1 
and  X  =  5. 

12.  Find  the  mean  ordinate  oi  y  =  x^  between  x  =  0  and  x  =  S, 
also  between  x  =  —  3  and  a;  =  +  3.    Why  are  these  results  the  same? 

13.  Find  the  mean  value  with  respect  to  x  of  the  square  of  the  or- 
dinate of  a  semicircle  of  radius  a. 

14.  Find  the  mean  sectional  area  of  a  sphere  supposed  cut  by  a  series 
of  equidistant  parallel  planes.    Explain  the  result  geometrically. 

15.  Find  the  mean  sectional  area  of  a  cone  of  radius  r  and  height  h 
supposed  cut  by  a  series  of  planes  parallel  to  the  base. 

16.  If  a  body  falls  vertically  from  rest,  its  velocity  v  at  the  end  of  t 
seconds  is  given  by  the  equation  v  =  32  /.  Find  the  average  velocity, 
(a)  for  the  first  second,  (b)  for  the  first  six  seconds  of  its  motion. 

17.  A  quantity  of  steam  expands  so  as  to  follow  the  law  pv°'^  =  200, 
p  being  the  pressure  measured  in  lbs.  per  sq.  in.  Find  the  average 
pressure  between  the  volumes  1  cu.  ft.  and  30  cu.  ft. 

18.  A  particle  moves  along  the  axis  of  x  so  that  the  force  upon  it 
at  a  distance  x  from  the  origin  is  equal  to  ax,  where  a  is  a  constant. 
Find  the  mean  value  of  the  force  as  x  increases  from  0  to  s. 

19.  If  a  body  moves  so  that  its  speed  is  v  =  t  +  -,  calculate  the 


INTEGRATION  121 

tlisUiiice  traversed  between  the  iiincs  t  =  2  :iii(l  t  =  4  and  the  average 
speed. 

20.  Find  the  mean  value  of  sin  x  between  ^  =  J  and  ^J  =  ^. 

21.  Find  the  mean  value  oiy  =  —  between  x  =  I  and  a;  =  4. 

22.  A  spring  oscillates  so  that  the  force  of  F  lbs.  which  it  exerts  on 
a  weight  at  the  end  of  t  seconds  is  given  by  /^  =  2  sin  3  t.  Find  the 
mean  value  of  the  force  from  i  =  0  to  ^  =  ^. 

23.  The  electric  current  C  in  a  conductor  at  time  t  is  given  by  the 
equation  C  =  4  sin  200 1.  Find  the  mean  value  of  C  from  <  =  0  to 
t  =  ^--  seconds. 

29.  Work,  Attraction,  Mass.  The  work  done  by  (or 
against)  a  force  is  defined  as  the  product  of  the  force  by 
the  distance  through  which  it  acts.  This  definition  can 
hold,  however,  only  when  the  force  is  constant,  unless  we 
take  the  distance  so  short  that  the  force  can  be  considered 
as  constant  over  that  distance.  We  thus  obtain  the  element 
of  work,  and  })y  summing  these  elements,  integratinn-.  over 
the  whole  path  we  obtain  the  total  work  done. 

Example  1.  A  coiled  spring,  two  feet  in  length,  resists 
compression  with  a  force  which  varies  as  the  distance 
through  which  the  end  of  the  spring  is  moved  in  compress- 
ing it.  How  much  work  is  done  in  compressing  the  spring 
to  half  its  original  length? 

Let  X  be  the  distance  through  which  the  end  of  the  spring 
is  compressed.  Then  the  force,  F  =  K  Xy  and  the  work 
done  in  compressing  it  a  very  small  distance,  dx^  is  the 
element  of  work  d  E  =  Kx  dx.    Therefore, 

E  =  Work  =   /    Kx  dx  =  -^'1  =  ^ 

•^0  2    Jo       2 

The  force  of  attraction  between  two  masses  is  defined  as 
the  product  of  the  masses  divided  by  the  square  of  the 


122  CALCULUS  AND  GRAPHS 

distance  between  them.     This  definition  holds  only  when  \ 
each  mass  is  concentrated  at  a  single  point  and  the  distance  ■ 
does  not  vary.     If  the  mass  is  that  of  a  body  having  di- 
mensions, we  must  take  a  portion  of  the  body  so  small 
that  the  whole  portion  can  be  considered  to  lie  at  a  con-  : 
stant  distance  from  the  mass  it  is  attracting.     We  thus ; 
form  the  element  of  attraction,  the  element  of  the  thing  j 
to  be  summed,   and  by  integrating  we  obtain  the  total 
force. 

Example  2,  Opposite  the  middle  point  of  a  very  thin  wire 
of  length  I  cm.  at  a  distance  c  centimeters  from  the  wire,  is  I 
situated  a  particle  of  unit  mass.    Find  the  force  with  which  ' 
the  wire  attracts  the  particle.    See  Fig. 

P  Let  X  be  the  distance 

-2 — tH —  from  the  center  of  the  \ 


1 T-  ,  .  . 

"2"  I  /  wire  to  any  point  P.   Let  | 

C\           /^  the  density  of  the  wirej 

1^^^  /  (mass  per   unit   length) ' 

\/^  be  p,  a  constant.    Then 

[4  the  element  of   mass  at! 

^'^'^^■^  P  is  p  dx  and  the  ele-l 

ment  of  force  of  attraction  between   p  dx    and  the  par-i 

tide  at  A  is  \ 

I  .  pdx  \ 

But  this  attraction  is  made  up  of  two  components  one 
parallel  and  one  perpendicular  to  the  wire.  The  component 
parallel  to  the  wire  is  neutralized  by  the  pull  of  another  i 
little  element,  p  dx^  at  the  same  distance  from  0  as  is  P  i 
but  on  the  other  side,  so  that  only  the  component  per-; 
pendicular  to  the  wire  is  effective.  If  we  use  OAP  as  thej 
triangle  of  forces,  AP  will  be  the  total  force  and  AO  the! 
component  at  right  angles  to  the  wire;  the  normal  com- ; 


INTEGRATION 


123 


ponent.    Therefore  we  have  for  the  normal  attraction  since 
AO  =  APcos^, 


dF'  =  dF  .  co^  6  = 


p  cos  d  dx 


and 


r 


/p  cos  d  dx^ 


This  integral  involves  three  variables,  6,  x  and  S.  Let 
us  express  it  all  in  terms  of  the  variable  d.  Thus,  since 
X  =  c  tan  d,  must  dx  =  c  sec^  Odd;  and,  since  S  = 
c  sec  d,  must  S^  =  c^  sec^  6.    Therefore, 


F'  = 


I 


2c p  cos  d 


.  csoc^ddd 


0       c^  sec^  0 

for  the  normal  attraction  of  half  the  wire.  It  is  obvious 
that  d  begins  with  the  value  zero,  and  ends  with  the 
angle  formed  by  joining  the  end  of  the  wire  to  A,     The 

I     I  I 

tangent  of  this  angle  is  2  =  — ,  so   that    6  =  tan  ^   —  • 

—        ji  c  ^  c 

c 

We  then  have  for  the  total  normal  force, 

tan-^J. 

9  r^ 

sin  6 


C    ^^  0  c 

=  — -  ]  sin  ( tan~^  -—  )  —  sin  0 


tan  _i. 
2c 


I 


2p  

c    •  ^p-\-4i 
2pl 


cVf  +  4  c^ 

M 
If  the  total  mass  of  the  wire  be  M,  then  9  =  ^1  ^^d  we 

may  write 


2F'  = 


M 

c\/P  +  4  c2 


124  CALCULUS  AND  GRAPHS 

Example  3.  Suppose  the  density  of  the  wire  in  Example  2 

is  not  constant  but  equal  to  -  —  x^  x  being,  as  before,  the 

distance  from  the  center  of  the  wire.     What  is  the  total 
mass  of  the  wire? 

In  this  case  the  element  of  mass,  at  P,  is 

dM  =  p  dx  =  \7y  —  x\  dx 
and 


^=/X^-^^)-=^/J6-^)^^ 


4 


EXAMPLES 

L  Under  a  circular  metal  plate  of  radius  20  centimeters  is  placed 
an  electro-ma,a;net  which  sets  up  a  field  of  force  in  the  plate  such  that 
the  force,  500  dynes  at  the  center,  is  always  equal  to  500  —  r^  where 
r  is  the  distance  from  the  center.  Find  the  work  done  in  pushing  a 
particle  from  the  center  to  the  edge  of  the  plate. 

2.  A  very  thin  wire  of  length  100  cm.  and  density  0.1  grammes  at- 
tracts a  unit  particle  in  the  line  of  the  wire  and  10  cm.  from  its  end. 
Find  the  force  of  attraction  between  the  wire  and  the  particle. 

3.  A  very  thin  homogeneous  wire  of  length  I  and  density  p  attracts 
a  unit  particle  in  the  line  of  the  wire  and  distant  S  units  from  its  end. 
Find  the  force  of  attraction  between  the  wire  and  the  particle. 

4.  Force  is  defined  as  the  product  of  mass  by  acceleration.  A  body  of 
unit  mass  moving  in  a  straight  line  has  an  acceleration  of  •\-  K^  x  ft. 
per  sec.  per  sec,  where  x  is  the  distance  from  a  fixed  point  0  and  K  is 
a  constant.  Find  the  work  done  as  the  body  moves  from  0  to  a  dis- 
tance C 

5.  The  weight  of  a  body,  which  is  a  force,  is  inversely  proportional 
to  the  square  of  its  distance  from  the  center  of  the  earth.  Find  the 
work  done  in  lifting  a  weight  of  10  lbs.  from  the  surface  of  the  earth 
to  a  height  of  one  mile  above  the  surface.  (Take  the  radius  of  the 
earth  as  4000  miles) . 

6.  Express  as  an  integral  the  work  done  against  attraction  by  moving 


INTEGRATION  125 

the  particle  of  Example  2  from  the  end  of  the  wire  to  its  position  10  i 
cm.  away.    (Use  the  result  of  Example  3.)  | 

7.  Express  as  an  integral  the  work  done  in  moving  the  particle  of  \ 
illustrative  Example  2,  Art.  29,  from  the  middle  point  of  the  wire,  0,  I 
to  the  position  A . 

8.  Find  the  total  mass  of  a  thin  circular  plate  of  radius  a  if  the  density  : 
at  a  distance  x  from  the  center  is  equal  to  o  —  x.    (Suggestion:  let  the 
element  of  mass  be  that  of  a  very  narrow  ring  at  any  distance  from  the 
center.)  i 

9.  Find  the  mean  value  of  the  density  of  the  plate  of  Example  8,  \ 
with  respect  to  distances  from  the  center.  Would  a  plate  of  constant ; 
density  equal  to  this  mean  have  a  greater  or  less  mass  than  the  plate  i 
of  Example  8?  ! 

10.  A  straight  line  bisects  the  angle  between  the  axes  of  x  and  y ' 
and  is  10  feet  long.    Find  the  mass  of  a  triangular  plate  made  by  the 
given  line,  the  axis  of  x  and  the  line  x  =  5\/2,  (a)  when  the  density  : 
varies  as  the  distance  from  OX;  (b)  when  the  density  varies  as  the  dis- 
tance from  OY, 


CHAPTER  VIII 
MISCELLANEOUS  EXAMPLES 

30.  Miscellaneous  Examples.  Given  the  function  y  =  (x  -\-  7) 
(x  —  2)  (x  —  4),  answer  questions  1-12. 

1.  Sketch  the  graph. 

2.  Plot  the  graph  accurately  from  x  =  —  7  to  x  =  4. 

3.  Find  maximum  and  minimum  values  of  the  function. 

4.  Find  the  direction  in  which  the  graph  crosses  OX;  OY. 

5.  Given  that  the  abscissa  increases  at  the  rate  of  2  feet  per  second, 
find  the  rate  at  which  the  ordinate  is  increasing  at  the  points  where 
the  graph  crosses  OX;  OY;  at  the  maximum  point;  at  the  minimum 
point. 

6.  Find  the  mean  value  of  the  ordinate  of  the  curve  between  x  =  —  7 
and  X  =  2;  between  x  =  2  and  x  =  4;  between  x  =  —  7  and  x  =  4. 

7.  Is  the  direction  of  the  curve  ever  northeast;  that  is,  at  45°  with 
OXf    Where? 

8.  Find  the  area  between  the  curve  and  the  axis  of  x  from  x  =  —  7 
to  x  =  4. 

9.  Find  the  abscissa  of  the  center  of  gravity  of  the  area  between  the 
curve  and  OX  from  x  =  —  7  to  a;  =  2. 

10.  Find  the  abscissa  of  the  center  of  gravity  of  the  area  between 
the  curve  and  OX  from  x  =  2  to  x  =  4. 

11.  Find  the  abscissa  of  the  center  of  gravity  of  the  area  bounded 
by  the  curve,  the  axis  of  x  and  the  axis  of  y. 

12.  The  abscissa  of  a  point  on  the  curve  is  measured  as  6.  Find  the 
value  of  the  corresponding  ordinate  and  compute  the  approximate 
error  in  the  ordinate  due  to  a  possible  error  of  .02  in  the  abscissa. 

13.  Sketch  the  graphs  of  the  four  functions 

(a)  2/  =  X  (x  —  1)  (c)  y  =  X  {x  —  1)^ 

(b)  y  =  x{x^  1)2  (d)  y  =  x{x—  ly 

126 


MISCELLANEOUS  EXAMPLES  127 

What  conclusion  might  be  drawn  from  the  behavior  of  the  graphs  at 
the  point  where  a:  =  1? 

14.  Sketch  i  he  graph  of  the  function  y  — 


15.  Sketch  the  graph  of  the  function  y  = 


x  +  2 
(x— l)(a;  — 3). 
X  —  4 


16.  If  the  tank  of  Ex.  10,  Art.  28,  is  10  ft.  long,  4  ft.  high  and  5  ft. 
wide,  find  the  pressure  on  the  greatest  side  (a)  at  the  beginning  of  the 
operation;  (b)  at  the  end  of  5  minutes;  (c)  using  the  mean  value  of  the 
density. 

17.  In  Example  16  find  the  mean  value  of  the  pressure  during  the 
5  minutes.    Compare  the  result  with  result  (c)  of  that  example. 

18.  Find  the  mean  value  of  the  pressure  in  Example  16  with  respect 
to  the  density. 

19.  A  triangular  plate,  of  negligible  thickness,  is  of  the  form  of  the 
area  between  the  coordinate  axes  and  the  line  represented  by  y  = 

3  X 
3  —  — - .    The  density  of  any  particle  of  the  plate  is  equal  to  the  dis- 
tance of  the  particle  from  OY.    Find  the  mass  of  the  plate. 

20.  A  tank  whose  length  is  25  and  height  10  is  kept  full  by  a  stream 
of  oil  so  that  the  density  is  decreasing  at  the  rate  of  2  units  per  minute. 
At  what  rate  is  the  pressure  on  the  face  (25  by  10)  changing? 

21.  In  a  triangle  we  have  given  the  angle  B  =  45°,  C  =  30°,  and 
the  side  a  =  100  ft.  Wliat  eiTor  is  caused  in  the  side  c  by  an  error  in 
a  of  0.2  ft.,  assuming  B  and  C  exact?    (Suggestion:  use  the  law  of  sines.) 

22.  Sketch  the  graph  of  the  function 

=  ^  +  1 

^       (x—l){x+2) 

23.  Sketch  the  graph  of  the  function 

_  3^—1 

^  "  ix  —  3){x~^) 

24.  A  steamer  has  cast  off  and  is  drifting  away  from  the  pier  with 
a  velocity  of  2  feet  per  second.  A  belated  passenger  jumps  in  a  straight 
line  from  the  dock  (on  a  level  with  the  steamer's  deck)  with  a  velocity 
of  b  feet  per  second,  after  the  steamer  has  drifted  5  feet  away.  The 
wind,  blowing  directly  from  the  steamer  to  the  man,  retards  him  wath 


128  CALCULUS  AND  GRAPHS 

a  resistance  of  a  feet  per  second  per  second.  How  long  will  the  pas- 
senger take  in  making  the  leap? 

25.  In  Example  24  if  a  =  10  ft.  sec.^,  and  6  =  12  ft.  sec,  will  the 
passenger  make  the  steamer?  In  what  time?  What  is  the  length  of 
his  leap? 

26.  In  Example  24  if  a  =  6  ft.  sec.^  and  6  =  10  ft.  sec,  in  what  time 
will  the  passenger  make  the  steamer?  With  what  length  of  leap?  Ex- 
plain your  answers  fully. 

27.  A  particle  is  moving  along  the  axis  of  x  according  to  the  law 
X  =  5  —  2t,  t  being  the  time.  Another  particle  moves  along  the  axis 
of  y  according  to  the  law  y  =4  —  St.  How  far  apart  are  the  particles 
when  they  start  to  move?  When  are  they  nearest  to  each  other,  and 
what  is  that  distance?    How  long  will  they  continue  to  approach? 

28.  A  particle  is  moving  on  the  axis  of  x  according  to  the  law  x  = 
cos  t;  another  particle  on  the  axis  of  y  according  to  the  law  y  =  sin  ^. 
When  are  they  nearest  to  each  other?  When  are  they  moving  with 
the  same  speed? 

29.  A  particle  is  moving  on  OX  according  to  the  law  a;  =  2  sin  t; 
another  on  OF  according  to  the  law  y  =  sin  2  ^.  When  are  they  near- 
est to  each  other?  When  farthest  apart?  What  is  the  least  distance? 
The  greatest? 

30.  Find  the  area  between  the  two  arches  above  OX  made  by  the 

4  3*       4  oT'^ 
curves  y  =  sinx  and  y  =  —  —  — • .     Plot  the  two  arches  in  one 

X         x^ 

diagram  on  the  same  scale. 

31.  A  parachute  falls  from  a  height  of  365.3  feet,  with  an  initial 
velocity  of  3  feet  per  second,  according  to  the  law  a  =  32  —  8  /.  In 
what  time  will  it  again  have  a  velocity  of  3  ft.  per  sec?  How  far  will 
it  then  have  dropped?    What  is  its  maximum  velocity? 

32.  A    belt    transmits    power,    P,    according    to    the    law    P  = 
T  — I  where  V  is  the  linear  velocity  of  the  belt  in  meters  per 

second,  W  is  the  weight  per  meter  in  kilograms  of  the  belt,  and  T  is 
the  tension  in  the  belt  while  at  rest.  What  value  of  Y  will  make  P  a 
maximum?  Find  the  maximum  power  when  IF  =  10,  T  =  2.  Note: 
g  =  9.81  meters  per  sec  per  sec. 

33.  If,  in  Example  32,  the  machinery  is  stopped  so  that  Y  decreases 
at  the  rate  of  a  meters  per  sec,  what  will  be  the  rate  of  change  of  F'^ 


MISCELLANEOUS  EXAMPLES  129 

34.  In  deep  water  the  velocity  of  a  wave  of  length  /  is  given  by  the  i 

/I       a 
function  F  =  i/-  +7, a  being  a  known  constant.    What  wave  length  < 
\  a       I 

will  make  the  velocity  a  minimum?  I 

35.  The  waste  due  to  heat,  depreciation,  etc.,  in  an  electric  conductor 

is  given  by  the  function  u  ==  c^R  +  — -,  where  R  is  the  resistance  in 

R 

ohms  per  mile,  and  c  the  current  in  amperes.     If  c  is  kept  constant 

what  value  of  R  will  make  the  waste  least?  j 

36.  The  gate  at  a  grade  crossing  has  an  arm  15  feet  long  over  the 
road  and  an  arm  6  feet  long  over  the  sidewalk,  the  arms  rotating  up>- 
ward  on  the  same  axis  at  the  rate  of  3  radians  per  minute.  At  what 
rate  is  the  distance  between  the  ends  of  the  arms  changing  when  they 
have  rotated  through  45°  from  the  horizontal?    Through  60''? 

37.  The  rate  at  which  the  surface  of  a  body  of  water  falls  is  given  by 

^  s_V — g  Kd      X)  ^jjgj.g  ^  jg  ^Yie  original  depth  of  the  water,  s  is 
d^  S 

the  area  of  the  cross-section  of  the  opening  through  which  the  water 
is  flowing,  and  S  is  the  surface  of  the  water.    How  long  will  it  take  to 
empty  a  vertical  cylindrical  tank  15  feet  high,  of  radius  3  feet,  through 
a  hole  of  1  inch  radius  in  the  bottom  of  the  tank?    How  long  will  it  1 
take  to  lower  the  water  3  feet  from  a  full  tank?     (Note:  take  ^  =  321 
and  t  in  seconds.) 

38.  The  acceleration  of  a  particle  starting  at  a  distance  h  from  the  ! 

center  of  the  earth  and  falling  toward  the  earth  is  given  by  the  function  i 

gR^  I 

/  =  — — ,  in  which  R  is  the  radius  of  the  earth,  g  the  gravitational 

{h  —  sr  I 

constant  and  S  the  distance  through  which  the  body  has  fallen.    Find 
the  velocity  of  the  particle  after  it  has  fallen  a  distance  S  (a)  from 

/,  X     .  ,   .  .  .  1      1     .  /*  d^      d^    ^^'»"  d^\ 

rest;  (b)  with  imtial  velocity  Vq.    I  Note:/  =  t=TT  =  *'tI 

\  at       its     (It         (is  J 

39.  In  Example  38  find  the  velocity  with  which  the  particle  reaches 
the  earth,  if  it  starts  from  rest  (a)  at  a  distance  2  R  from  the  center  of 
the  earth;  (b)  at  an  infinite  distance.  (Suggestion:  In  the  result  (a)  of 
Example  38  put  s  =  h  —  R.) 

40.  Taking  R  =  4000  miles  and  ^  =  32  feet  sec. 2,  compute  the  re-  I 
suits  of  Example  39.  J 


130  CALCULUS  AND  GRAPHS 

41.  A  curve  is  given  by  the  equations  x  =  f^  and  y  =  4  i  —  t^j  where 
t  is  the  time  in  seconds,  x  and  y  being  in  feet.  Sketch  the  curve.  [Sug- 
gestion: Give  values  to  t  and  plot  the  points  (x^  y)]. 

42.  Find  the  area  of  the  loop  of  the  curve  in  Example  41.  (Sugges- 
tion: Express  the  integral  in  terms  of  ^.) 

43.  Find  the  volume  got  by  revolving  the  loop  of  the  curve  in  Ex- 
ample 41  about  the  axis  of  x.  (Suggestion:  Express  the  integral  in 
terms  of  ^.) 

44.  In  what  direction  is  the  particle  which  generates  the  curve  of 
example  41  moving  at  the  end  of  2  seconds? 

45.  At  what  angle  does  the  curve  of  Example  41  cross  the  axis  of  x? 

46.  What  is  the  speed  of  the  particle  in  Example  41  when  it  starts 
to  move?  When  it  crosses  the  axis  of  x  for  the  first  time?  At  the 
highest  point  of  the  loop? 

47.  When  and  where  will  the  components  of  velocity  of  the  particle 
in  Example  41  be  equal? 

48.  On  the  ordinate  of  a  point  on  the  circle  x"^  -\-  y-  =  a^,  at  right 
angles  to  the  plane  of  the  circle,  a  rectangle  is  constructed  having  the 
ordinate  in  question  as  base  and  the  abscissa  of  the  point  as  altitude. 
Find  the  mean  value  of  the  area  of  the  rectangle  as  x  passes  from  o  to  a. 

49.  Given  a  body  moving  according  to  the  law «;  =  32  ^  (^  in  seconds), 
find  (a)  the  distance  traveled  in  the  first  2^  seconds;  (b)  the  average 
value  of  the  velocity  during  the  time  named. 

50.  Given  a  body  moving  according  to  the  law  v  =  \/2  gS,  find  the 
average  value  of  the  velocity  during  the  first  hundred  feet.  Compare 
the  result  with  the  result  of  Example  49. 

51.  Sketch  the  graphs  of  the  three  fimctions 

(a)  y  =  (a:— l)M^-f2) 

(b)  2/  =  (x—iy{x+2) 

(c)  y  =  {x—irix+2) 

52.  Plot  the  graph  of  each  of  the  functions  in  Example  51  at  inter- 
vals A  X  =  .1  from  a;  =  .5  to  a;  =  1.5. 

53.  Find  the  direction  of  each  of  the  graphs  of  Example  51  at  (a) 
a;  =  1;  (b)  a;  =  1.1. 

54.  Find  the  area  under  the  arch  for  each  curve  in  Example  51. 

55.  If  each  of  the  graphs  in  Example  51  is  the  path  of  a  moving  par- 
ticle, and  if  the  y-component  of  velocity  in  each  of  the  three  is  2  feet 


MISCELLANEOUS  EXAMPLES  131: 

per  second,  what  is  the  speed  in  its  path  for  each  particle  at  the  point 
whose  abscissa  is  1  foot?    At  the  point  whose  abscissa  is  1.1  foot? 

56.  Gravity  acting  downward,  parallel  to  OY,  find  the  turning  mo-| 
ment  with  respect  to  OY,  due  to  the  force  of  gravity  acting  on  a  very 
thin  plate  in  the  shape  of  the  first  arch  of  the  curve  y  =  sin  x,  The| 
weight  of  any  portion  of  the  plate  is  proportional  to  the  reciprocal  of  | 
the  distance  of  that  portion  from  OF,  and  is  equal  to  ^  when  x  =  2.     j 

57.  The  edges  of  a  very  thin  plate  in  the  form  of  an  equilateral: 
triangle  have  for  their  equations  ^ 

A  hole  in  the  shape  of  an  equilateral  triangle  with  its  sides  parallel  tol 
the  sides  of  the  plate,  is  cut  out  of  the  center  of  the  plate.    The  equa-. 

VS  -  VS  -  i 

tions  of  the  sides  of  the  hole  are  y  =  —  —  .t-v/S,  2/  =  ~7^  +  ^VSi 

18  18  ' 

and  y  =  —  — .    If  the  foot  is  the  unit  of  length,  show  that  the  edged 

oU  I 

of  the  plate  are  4  feet  long;  the  edges  of  the  hole  2  inches  long.  Also, 
if  the  weight  of  the  material  of  the  plate,  in  ounces  per  square  foot,  ia! 
equal  to  the  distance  from  the  vertex  find  the  total  weight  of  the  platej 
(Suggestion:  Take  strips  of  area  parallel  to  OX.)  I 

58.  In  example  19  find  the  mean  value  of  the  density  of  the  plate.j 
If  the  plate  were  of  constant  density  equal  to  the  mean  what  would] 
be  its  mass?  I 

X  I 

59.  Sketch  the  graph  of  the  fimction  y  =  sin  a;  +  2  sin  -.     (Sugges^ 

X 

tion:  put  yi  =  sinx,  2/2  =  2  sin -,  plot  these  graphs  in  the  same  dia- 

^  '' 

gram  and  note  that  y  =  yi  -\-  2/2.)  ' 

60.  Sketch  the  graph  of  the  function  y  =  cos  x  —  sin  2  x*  \ 

61.  Sketch  the  graph  of  the  function  y  =  ^sinx  —  3^  sin  2  x.*  ' 

62.  Sketch  the  graph  of  the  function  y  =  ^  cos  2  x  -}-  |-  cos  3  x.* 

63.  Find  maximum  and  minimum  values  of  the  functions  given  in! 
Examples  59-62. 

64.  Find  the  area  under  the  first  arch  of  each  of  the  curves  given  in! 
Examples  59-61. 

*  See  suggestion  in  Example  59.  j 


CHAPTER  IX 

EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS 

31.  Exponential  Functions  and  their  Graphs.  An  ex- 
ponential function  is  one  in  which  a  variable  occurs  in  the 
exponent.  For  example,  y  =  e'^jy  =  a^^""  etc.  The  graphs 
of  such  functions  can  be  plotted  in  the  usual  way  by  giving 
values  to  the  independent  variable  (x)j  computing  the 
corresponding  values  of  the  function  (?/),  and  plotting  the 
points  thus  found. 


Example  1. 

y  =  ( 

f.*    SeeFi 

g.  28. 

X 

0 

1 

2 

3 

00 
00 

e 

—  1 

—  2 

—  3 

00 

y 

1 

e 

e'- 

e^ 

1 

e 

1 

e' 

1 

7 

1 

y 

1 

2.7 

7.3 

19.7 

oo 

.37 

.14 

.05 

0 

*  The  number  e,  the  base  of  the  Napierian  or  natural 
system  of  logarithms,  may  be  defined  as  the  limit  of  the 
sum  of  the  terms  of  the  following  series  as  the  number  of 
terms  increases  without  Umit. 


=  1+-^  + 


T  "1"r2   "^JS  "f  J4 


+.^  +  - 


=  2.7182818  . 


In  sketching  graphs  we  shall  use  the  approximate  value 
e=2.7. 


FiQ.  28 


-2       -1     O 


X 


132 


EXPONENTIAL  FUNCTIONS 
Example  2.  y  =  xe^~^ 


133 


X 

0 

1 

2 

3 

4 

n 

—  1 

—  2 

—  n 

y 

0 

1 

2 
e 

3 

4 

—  e' 

~2e' 

-^6"+' 

y 

0 

1 

.74 

.42 

.20 

0 

—  7.3 

—  39.4 

00 

When  X  =  n,  and  n  is  very  large,  e'^     becomes  very  large 

n 
even  as  compared  with  n,  so  that  the  fraction  -;^,  whose 

denominator  is  very  large  as  compared  with  its  numerator, 
becomes  very  small.     We  express  this  by  writing,  briefly, 

?/  =  -^  =  0.     Similarly,  when  x  =  — n  and  n  becomes, 
e 

numerically,  very  large,  we  write,  briefly,  y  =  —  ^  .  e^ 

=  —  00  .    See  Fig.  29. 


Fia.  29 


134 


CALCULUS  AND  GRAPHS 


Example  3.  y  =  e  ^  sinx 

Using,  as  in  Art.  3,  the  limiting  values  of  the  sine  we  put 


Sin  a:  =  0 

X   =   0,   TTj  2  TT  .  ,  ,  — 

^  =  0,  0,  0  ...  0,  0 


TT, 


•2  TT 


X   = 


Sin  X  =  —  1 

3  TT    7  TT 

~2~'~2~" 
1 


y  = 


7%   ' 


Sin  X  =  1 

TT     5  TT 

J_        1 

y  =  iz'  ~5i 

z,2  ^"2- 


Thus  the  graph  is  a  wave  curve  crossing  the  axis  of  x  an 
infinite  number  of  times  at  equal  intervals  (tt),  and  with 
the  crest  of  the  wave  lower  and  lower  as  x  increases  from 
zero.  For  negative  values  of  x  the  factor  e'^  is  positive  and 
increases  as  x  increases  numerically,  so  that  the  crest  of 
the  wave  rises  higher  and  higher  as  x  passes  from  0  to  —  oo  . 
See  Fig.  30. 
Y 


Fig.  30 


LOGARITHMIC  FUNCTIONS 


135 


32.  Logarithmic  Functions  and  their  Graphs.  A  logarith- 
mic function  is  one  which  involves  the  logarithm  of  a  va- 
riable. For  example,  y  =  loggo:,  y  =  logjo  a/1 — a:^,  etc. 
We  can  plot  such  functions  by  giving  values  to  the  inde- 
pendent variable  (x),  computing  or  finding  in  tables,  the 
corresponding  values  of  the  function  {y)  and  plotting  the 
points  thus  obtained.  It  is  usually  simpler,  however,  par- 
ticularly in  sketching  the  graphs  of  logarithmic  functions, 
to  treat  them  as  the  inverse  of  exponential  functions  and 
to  proceed  as  in  Art.  31. 

Thus,  given,  Example  1,  the  function  y  =  logio^:,*  we 
know,  by  the  definition  of  a  logarithm,  that  x  =  10^,  which 
can  be  plotted  by  the  methods  of  Art.  31.  Thus  (see 
Fig.  31). 


y 

0 

1 

2 

00 

—  1 

—  2 

00 

X 

1 

10 

100 

00 

.1 

.01 

0 

Fio.  31 

*  Horcaftrr  wlion  no  base  ia  inontiontHl  the  Napierian  base,  e,  will  bo  understood. 


136  CALCULUS  AND  GRAPHS 

EXAMPLES 

Sketch  the  graphs  of  the  following  functions: 


1. 

y=e-^' 

2. 

2/  =  gVx 

3. 

y 

^  xe^ 

4. 

2/  =  a:  e-^ 

5. 

y  =  X  el^ 

6. 

y 

=  x2  e-^ 

7. 

y  =  ei-^ 

8. 

y=^(e--^e-n 

9. 

y 

=  ^{e--e--) 

10. 

y  =  e~^  cos  X 

11. 

y  =  e~^  sin  2  x 

12. 

y 

=  e~^^  cos  x 

13. 

y  =  log2  X 

14. 

2/  =  log3  X 

15. 

y 

=  10g4  X 

16. 

y  =  logio  sin  x 

17. 

y  =  logio  cos  X 

18. 

y 

=  logio  tan  X 

19. 

1      1-^ 

20. 

y  =  logio  {x  +  1) 

21. 

y 

=  log  ix  +  1) 

22. 

23.  2/  =  K 

^-h^x^ 

24.  Plot,  in  one  diagram,  the  two  curves 

(6—1)2 


2/  =  |-  (e^  +  e  ^)  and  y—l  = 


2e 


'  x^ 


33.  Derivatives  and  Integrals  of  Exponential  and  Log-  j 
arithmic  Functions.    We  notice  in  the  case  of  the  second  \ 
function  of  Art.  31  (Fig.  29)  that  it  rises  to  some  highest  i 
point  and  then  descends;  that  the  function  has  a  maximum  ; 
value.    We  have  learned  (Art.  15)  how  to  find  maximum  I 
and  minimum  values  of  a  function  by  means  of  the  deriva-  | 
tive.     Also  in  Chapters  III  to  VII  we  have  learned  to  ; 
apply  the  derivative  and  the  integral  to  many  problems 
connected  with  functions.     All  these  applications  can  be 
made  to  exponential  and  logarithmic  functions  provided 
we  can  find  their  derivatives  and  integrals.    This  we  shall  j 
now  proceed  to  do.  i 

Let  y  =  log  X  \ 

Then  A  ?/  =  log  (oj  +  A  x)  —  log  x  \ 

=  iog-_+^  =  iog(n-Af)  I 

X  \  X     /  j 


LOGARITHMIC  FUNCTIONS  137 

and  .      /         Ax\  i 

Ax  Ax  \  X  / 


which  may  be  written 


1_  _£_ 

A  a:\A^ 

nf        lr\(r  I     I    -I I  innr  i     i    -*-_ 

Ay 
A  X 


Therefore 


limit  A^  _  ^ 

Ax  =  o^  a; 


limit  log /^^   I    Arr\A^ 


log  limit  A    ,    A^y^ 


X 

X 

(A  x\a^ 
1  +— —  j      =  e*  and  log  e  =  1 


Therefore 


d^  ^d  logx  ^  1  (52) 


*  Let  ^-^  =  /i,  then  -—-  =  t  and 
X  ilx       h 


limit        ( 1  +  ^)  A x  =  limit    ^  ^  j^^h  ^  [expanding  by  the  binomial  theorem]  ] 

^x  =  0  ^  X   ^  n  =  u 


^[  h  \2  13  J 


in 

limit 
h 


=  limit    L     ,    1    ,l-h(l-h)  (l-2/i)  \ 

;»=0|^  "*"l   +     |2_      +  |3_  +1 


138  CALCULUS  AND  GRAPHS 

Remembering  that  integration  is  the  reverse  process  of  dif- 
ferentiation, it  is  obvious  from  (52)  that 


/dx       , 
—  =  log  X 
X 


If  y  =  log  u  we  see,  by  foot-note  Art.  13,  page  41,  and  by 
(52)  that 

dy  _  dlogu  _  dlogu     du  _  1     du  (54) 

dx  dx  du      '  dx       u  '  dx 

li  y  =  log^  u  we  may  write 

l02!    U  * 

y  =  log«  u  =  p-^1— 

Then 

dy  _      1        dlogu  _      1         1     du 
dx       log  a  '      dx  log  a  '  u  '  dx 

or 

d  loga  u  _        1  du  _  loga  6     du  (55) 


(56) 


dx           uloga  '  dx           u     '  dx 

In  particular 

d  log,  x  _       1       _  loga  e 

dx           X  log  a           x 

Consider  next  the  function  y  =  e"" 

Then 

u  =  logy 

and 

du  _  dlogy  _  d  log  y     dy       1     dy 
dx           dx              dy      '  dx       y  '  dx 

or 

dy  _     du 
dx  ~  ^  dx 

*  Passano'a  Trigonomptry,  Art.  32. 

LOGARITHMIC  FUNCTIONS  139 

That  is      de"  _   ^du         a^''  -    '^  (5'^) 

~ —  —  6    ~;~«      3(11(1      ;      —  6  • 

dx  ax  ax 

It  is  obvious  that 

Je"  dx  =  e"  (58) 

If  y  =  ci^  we  may  write 

log  y  =  u  log  a 

d  log  y  _  1        du 
dx  dx 

1     dy  _  .         du 
y  '  dx  dx 

d^  __     ,         dM 
dx  dx 

or  da^  —    w  1         ^  (^^) 

dx  dx 

Also  da^        ^,  (60) 

-r~  =  a  log  a  ^    ^ 

dx 

It  is  obvious  that 

I  a"^  log  a  dx  =  log  a   j  a"^  dx  =    j  da^  =  (f 

^  log  a 

For  convenience  of  reference  formulae  52-61  are  here  as- 
sembled. 

d  logu  _  1     ^  dlogx  _  1 

dx  u  '  dx  dx  X 

d  logg  ti  _       1  du         d  log,  X  _       1 

dx  u  log  a  *  dx  dx  x  log  a 


140 


CALCULUS  AND  GRAPHS 

de"  _    udu 
dx           dx 

dx 

(62) 

dd^         u  1         du 

da""         xi 
dx  ='^^°^'^ 

f  X    =^^^^'     f'^ 

dx 

=  e"",      J   a""  dx  = 

a' 

[a 

34.  Integration   by   Parts.      In   addition   to   the   three  | 

methods  of  integration  given  in  Art.  18  there  is  a  fourth,  ! 

integration  by  parts,  which  is  particularly  useful  in  dealing  ■ 
with  exponential  and  logarithmic  functions. 

We  have  proved,  Art.  8,  that  | 

i 

duv         du   .       dv  \ 

~T~  =  ^T  +  ^y  ' 

dx          dx          dx  \ 


or 


where 


or 


duv  f   .        f 

— —  —  'ou  -\-  uv 
dx 


,       du  ,        ,       dv 

u    =  -r  and  v    =  -r 

dx  dx 

j   u^  dx  =  u  and  I    v'  dx  =  v 


We  may  write,  therefore, 

d  (uv)  =  vu^  dx  +  uv^  dx 
or 

vu^  dx  =  d  (uv)  —  uv^  dx 
Then 


I    vu^  dx  =  uv  —   I    uv^  dx 


(63) 


If  then  we  have  an  integrand  consisting  of  the  product  of 
two  factors  (vu^)  one  of  which  (u^)  can  be  integrated  sep- 


LOGARITHMIC  FUNCTIONS  141 

arately,  we  can  replace  the  integral  of  such  an  integrand 
by  an  expression  of  the  form  uvj  minus  a  new  integral 
which  may  be  capable  of  integration  by  some  of  the  proc- 
esses already  learned. 

Formula  (63)  can  best  be  learned  in  words.  Thus: 
The  integral  of  the  product  of  two  factors,  one  of  which 
can  be  integrated  separately,  equals  the  product  of  the 
integral  of  that  one  factor,  times  the  other;  minus  a  new 
integral  consisting  of  the  same  integral  of  the  one  factor 
times  the  derivative  of  the  other. 

The  following  mnemonic  device  may  help  in  remember- 
ing the  theorem: 


/ 


(one    factor)    (other)  =  (/  one)    (other) 
u'  V  u  V 


I 


(/  one)  (derivative  other)         (64) 
u  v' 


e^  dx 


Example  L 

/xe^dx  =    r  e^   xdx   =  e^  .  X    —    C 
J  (one)  (other)     (/one)  (other)  ^ 

=  xe""  —  e^  =  e^  {x  —  1) 
Example  2. 

j  log  X  dx  =   11.  log  X  dx   =  x  log  X  —  /   ^  •      -  dx 

(one)  (other)      (/  one)  (other)      (/one)  (deriv.  other) 
=  X  log  X  —  x  =  X  (log  X  —  1) 

Example  3, 
I  x^  cos xdx  =  x^s\nx  —  3  I  x'^  sin  x  dx 
(integrate  again  by  parts) 

=  x^  sin  X  —  3  j  —  x^  cos  a:  +  2   /  x  cos  x  dx  [ 


142  CALCULUS  AND  GRAPHS 


=  a;'sinx  +  3x^cosa; — 6 


-J  sir 


X  sin  X  —  I  sin  x  dx 


=  rr^  sin  a;  +  3  x^  cos  x  —  6  x  sin  a;  —  6  cos  x  \ 

=  {x^  —  6  x)  sin  a:  +  (3  x'^  —  6)  cos  x 

Example  4.  i 

/  r  -1  -1  r_xdx  I 

cos    X  dx  =    j    1  .  cos    X  dx  =  X  cos    x  -\-   J      /-. -^  \ 


=  X  cos  ^  X  —  \/l 


■X 


,2 


Example  5.  Find  maximum  and  minimum  values  of  the 
function  y  =  xe^-\    (Ex.  2,  Art.  31.) 

dy  __    i_x  dx   .       de^'"" 
dx  dx  dx 

dx 

=  e^'^  —  xe^-'  =  e^-'  (1  —  x)  =  0 


x 
dy 
dx 

y  =  incr.      1  deer. 

Max. 

That  is,  the  function  has  a  maximum  value,  1,  when  x  =  1. 
This  value  would  help  us  to  sketch  the  graph  in  Fig.  29. 

Example  6.  At  what  angle  does  the  curve  y  =  xe^'""  cross 

the  axis  of  x?     As  in  Example  5,  ^  =  e^~^  (1  —  x)^  which 

ax 


e'-^  = 

0 

1  — 

X  =  0 

X  = 

00 

X  =  1 

= 

1 

1 

2 

^  = 

+ 

0 

- 

= 

incr. 

1 
Max. 

d 

LOGARITHMIC   FUNCTIONS  143 

when  y  =  0  and,  therefore,  x  =  0,  has  the  value  e.    There- 
fore the  curve  crosses  ox  at  tan"^  e  =  69°  48'. 

Example  7.  Find  the  area  between  the  curve  y  =  xe^'"' 
from  X  =  0  to  X  =  oo  . 

xe^'""  dx  =    /     X  .  e  .  e'""  dx  =  e    j   xe'""  dx 
Integrating  by  parts 

/  xe'""  dx  =  —  xe""  +  j  e'""  dx  =  —  xe'""  —  e~''  =  —  ^^"^  ^ 
Therefore 

Curve-area  =  e  |       ^  ^^1    =  c  (0  +  1)  =  e 

Example  8,  Find   the   mean   value  of    logio  x  between 
X  =  1  and  x  =  10. 


M.  V.  = 


/lO 
log 


10 


10  —  1 
Integrating  by  parts 

/    1  .  logio  X  dx  =  X  logio  X  —   /  ^  •  — ^^  ^^ 

=  X  logio  ^  —  ^  logio  e 
Therefore 

-110 

X  (logio  X  —  logio  e) 


M.  V. 


9 

10  (1  —  logio  e)  +  logio  e 
9 

^^-l^^^^^'  =  0.6768 


144  CALCULUS  AND  GRAPHS 

Example  9,  Find  the  rate  of  increase  of  the  logarithm  of 
a  number  as  compared  with  the  number,  for  the  base  e, 
the  base  10,  the  base  3.  When  the  number  is  greater  than 
unity,  does  the  logarithm  increase  faster  or  slower  than 
the  number? 

Let    y  =  loga  X 

dy  _       1       _  log,  e 
dx      xloga  X 

dy  _1 


V,  Base  6.      , 

dx       X 

2°.  Base  10.  J  =  i^i^'  =  ^^^^ 
dx  X  X 

3°.  Base  3.     J  =  -r^  =  77^— 
dx      X  log  3       1.0986  x 


When  a;  >  1  the 
logarithm  increases 
slower  than  the 
number. 


Example  10.    C  x   -        1 
I    e  sm  X  dx 

Integrate  by  parts. 

/e"^      sin  X  dx  =  e""  sinx  —   P  e^  cos  x  dx  (a) 

(one)  (other)  J 

The  new  integral  thus  got  is  not  simpler  than  the  first.    Let 
us  integrate  using  sin  x  as  the  factor  to  be  first  integrated. 

/e^      sin  X  dx  =  —  e"^  cos  x  -\-    C  e^  cos  x  dx  (b) 

(other)  (one)  J 

We  now  notice  that  (a)  and  (b)  are  two  equations  which 
can  be  solved  for    /    e^  sin  x  dx  by  adding  (a)  and  (b),  and 

for  I    e""  cos  x  dx  by  subtracting  (a)  from  (b).     Thus 


LOGARITHMIC  FUNCTIONS  14Sl 

I   e'^  sin  X  dx  =  e'^  smx  —    I  e*  cos  x  dx 

Adding  j 

2   16^  sin  X  dx  =  e^  (sin  x  —  cos  x)  j 

/ 

Subtracting 


X    -        J,        e"^  (sin  X  —  cos  x) 
6  sin  X  ax  =  — ^ 


0  =  —  c""  (sin  X  +  cos  x)  -\-  2   I  e^  cos  x  dx 


x)  +  2fe 


or 


/j;            7         ^"^  (sin  X  +  cos  x) 
e   cos  X  ax  = i 

35.  The  Compound  Interest  Law.  This  law,  also  known 
as  the  law  of  organic  growth,  and  the  simple  mass  law,  states 
the  fact  that  a  variable  quantity,  y^  has  a  rate  of  increase 
with  respect  to  an  independent  variable,  x,  which  is  pro-j 

portion al  to  the  quantity  (y)  itself.  | 

1 

Thus  ,  g  =  2^,  I 

where  i^  is  a  constant.  I 

We  may  write  1 

^=  Kdx    and     f^=K  f  dx  \ 

y  ^    y         ^ 

whence 

log  y  =  Kx+  ci 


146  CALCULUS  AND  GRAPHS 

and  we  see  that  the  function  or  quantity  whose  rate  of 
change  is  proportional  to  itself  is  the  exponential  function 
ce   . 

Example  L  How  much  would  $100  amount  to  in  10 
years,  at  6  per  cent  per  annum,  if  the  interest  were  com- 
pounded each  instant?  * 

Here 

dt  y 

log  y  =  mt-{-  c     or     y  =  ce-^^  ^ 
But  when  t  =  0,  y  =  $100     .  * .  c  =  100  and 

y  =  100  6-^^' 
When     ^  =  10    this  becomes 

y  =  100  e-^    =  $182.20 

Example  2.  Newton's  law  of  cooling  for  an  object  cooled 

in  moving  air  or  running  water  is  given  by  —  =  —  KQ 

where  i  is  the  time  and  Q  the  difference  in  temperature  be- 
tween the  object  and  the  fluid.     An  object  at  100°  centi- 

grade  cools  for  1  hour  according  to  the  law  -77=  —  .5  0. 

What  is  its  temperature  at  the  end  of  the  hour? 

"^  ^  =  —  0.5  (9;     whence  Q  =  ce"^-^' 


di 

when  ^  =  0,  0  =  100  .  ' .  c  =  100  and  0  =  100  6"^  ^^ 
When  i  =  1,  Q  =  60.6°. 

*  Of  course  in  actual  practice  interest  is  compounded  at  some  definite  period  such 
35  6  months  or  a  year. 


LOGARITHMIC  FUNCTIONS  147 

How  long  will  it  take  the  body  to  cool  from  lOO"*  to  10°? 
d  =  100  e"^  ^',     0  =  10  gives    10  =  100  e'^'^  or 

e^'-''  =  10     .  • .  0.5  ^  logio  e  =  logio  10  =  1 

^  =  ;r^l =  4  hrs.  36  min. 

0.5  logio  e 

Or  we  might  proceed  thus 

-110 

i  =  —  21ogl9       =—2  (log  10  — log  100)  =2  log  10  =  4.606 

J 100 

=  4  hrs.  36  min. 

EXAMPLES 

Find  maximum  and  minimum  values  of  the  following  functions: 

1 

I.  y  =  e~^  2.  y  =  e^  S.  y  =  X(F 

1 
4.  y  =  xe~^  5.  2/  =  ^(^  6.  2/  =  x^e~^ 

7.  2/  =  el-^  8.  2/  =i(e^+e-^)  9.  7/ =  |- (e^  —  e"^) 

10.  Find  the  area  under  the  first  and  second  complete  arches  of  the 
curve 

y  =  e~^  cos  X. 

11.  Find  the  area  under  the  first  and  second  complete  arches  of  the 
curve  y  =  e"^  sin  2  x. 

12.  Find  points  at  which  the  curves  y  =  e~^  sin  x  and  y  =  e~^  have 
the  same  slope. 

13.  Find  points  at  which  the  curves  y  =  e~^  cos  x  and  y  =  e~^  have 
the  same  slope. 

In  the  following  examples  x  is  the  distance  in  feet,  t  the  time  in  sec- 
onds, of  a  particle  moving  in  a  straight  line  according  to  the  law  given 
by  the  function.     Find  the  position,  velocity  and  acceleration  of  the 


148  CALCULUS  AND  GRAPHS 

particle  at  the  time  named;  when  the  velocity  will  be  zero;  the  maximum 
velocity. 

14.  a;  =  e  1 0  sin  2 1;    t  =  5. 

15.  X  =  5e~'cos^;    t  =  1. 

1 

16.  a;  =  2  e2  sin  2  /;    i  =  | 

17.  a;  =  4 e"* sin t;    i  =  | 

dy 
Find  — ■  for  the  following  functions: 
ax 

I X  1  +  a; 

18.  y  =  log  — - —  19.  y  =  log  ; • 

1  +  ^  1  — X 

20.  y  =  log  -^  21.  y  =  log  -^ 

X —  1  X  +  1 

22.  2/  =  log]  X  (x  +  1)  [  23.  2/  =  logV^'  +  2  x 

24.  Find  the  area  between  the  axis  of  x  and  the  curve  xy  =  5  from 
X  =  1  to  a;  =  5. 

25.  The  relation  connecting  the  pressure  and  volume  of  a  certain 
gas  is  expressed  by  pv^-^  =8.  Sketch  the  graph  of  the  function  and 
find  (a)  the  area  between  the  curve  and  the  axis  of  p  from  p  =  1  to 
p  =  4;  (b)  the  area  between  the  curve  and  the  axis  of  v  from  z;  =  1  to 
V  =  4. 

Integrate  the  following  expressions: 

26.  /2^c^a:  27.     /  5^^  rfx  28.     /  'K^'' dx 

29.     I^^^^dx  30.     I    dx  31.     l-^^dx 


32 
35 

38 


/'    ^x  C    dx  C 

/Vf^a;  36.     C_f^^_dx_^  C  dx 

e^  +  l  J2e2^  — 5  ^^'Jxloga: 

/cos  a;  c?a;  /*   sec^  x  dx  f*  si 

1  +  sin  a;  J  2  tan  x  -\-  S  J  co^ 


log  cos  X 

dx 
3x+2 

c?a; 
log: 

sin  2  X  dx 


cos  2  a;  +  4 


LOGARITHMIC  FUNCTIONS  149 

41.  Find  the  area  between  the  axis  of  x  and  the  curve  y  =  e^  —  e"* 
from  X  =  0  to  X  =  2. 

42.  Find  the  area  between  the  axis  of  x  and  the  curve  y  =  e'^,  sin  a; 
from  X  =  0  to  X  =  x. 

Integrate  the  following  expressions: 

43.  I  xf'^^  dx  44.     I  X  sin  xdx  45.    j  x  cos  2xdx 

46.    /  X-  sin  X  dx  47.     /  sin"^  x  dx  48.     I  tan~^  x  dx 

49.     je^ame^dx  50.     \  xe^^  dx  51.     j  x^  dx. 

Find  the  volume  got  by  revolving  about  the  axis  of  x  each  of  the 
following  curves: 

52.  y  =  e^  from  x  =  0  to  x  =  2. 

53.  y  =  e~^  from  x  =  0  to  x  =  oo  . 

54.  2/  =  e^  +  e~*  from  x  =  —  1  to  x  =  1. 

55.  2/  =  x*/'  e^  from  x  =  0  to  x  =  1. 

56.  A  particle  moves  in  a  straight  line  according  to  the  law  s  = 
a  (e^  +  e"0.  Find  (a)  the  velocity  and  acceleration  at  any  time  t; 
(b)  the  minimum  velocity. 

57.  The  acceleration  of  a  particle  moving  in  a  straight  line  is  given 

dh 
by  — -  =  a  (e'  +  e  0.     If  it  starts  from  rest  at  <  =  0,  how  far  will  it 
dt^ 

travel  in  the  first  5  seconds?    What  will  be  its  velocity  at  the  end  of 

the  5  seconds? 

58.  The  Napierian  logarithm  of  the  number  6.70  is  calculated  as 
1.90209.  If  there  is  an  error  of  0.0002  in  the  logarithm,  what  error  is 
caused  in  the  number? 

59.  The  number  corresponding  to  a  known  Napierian  logarithm, 
2.21047,  is  calculated  as  9.11.  If  there  is  an  error  of  0.01  in  the  number 
what  error  is  caused  in  the  logarithm? 

60.  The  common  logarithm  of  the  number  2564  is  calculated  as 
3.40890.  If  there  is  an  error  of  .00002  in  the  logarithm  what  error  is 
caiLsed  in  the  number? 


150  CALCULUS  AND  GRAPHS 

61.  The  increase  per  second  in  the  number  of  bacteria  in  a  cubic 
inch  of  culture  is  proportional  to  the  number  of  bacteria  present.  If 
A^  is  the  number  of  thousand  bacteria  per  cubic  inch  and  t  is  the  time 
in  seconds,  express  A^  as  a  function  of  t. 

62.  In  Example  61  how  long  would  it  take  the  bacteria  to  increase 
from  A^i  to  A^2  per  cubic  inch? 

63.  The  rate  of  decrease,  with  respect  to  the  height,  of  atmospheric 
pressure  above  the  earth's  surface  is  proportional  to  the  pressure.  Ex- 
press the  pressure  p  as  a  function  of  the  height,  h, 

64.  In  Example  63  if  the  pressure  is  762  mm.  at  sea  level,  express  p 
as  a  function  of  h;  also  if  p  =  30  in.  at  sea  level. 

d  fi 

65.  A  body  cools  in  moving  air  according  to  Newton's  law  —  = 

dt 

—  Kd.  lid  falls  from  40°  C.  to  30°  C.  in  200  seconds  what  is  the 
value  of  the  constant  Kf  How  long  will  it  take  for  Q  to  fall  from  30° 
to  20°?    From  20°  to  10°? 

66.  If  a  rotating  wheel  is  stopped  by  water  friction  the  rate  of  de- 

(dO)  \ 

— —  =  — Ko)  1. 
dt  J 

Express  the  angular  speed  as  a  function  of  the  time.  Find  the  value 
of  K  if  the  speed  of  the  wheel  decreases  50  per  cent  in  one  minute. 

67.  How  long  will  it  take  the  wheel  of  Example  66  to  slow  down 
from  100  revolutions  per  minute  to  10  revolutions  per  minute? 

68.  Find  the  mean  value  of  the  ordinate  of  the  curve  y  =  e^  from 
a:  =  0  to  a:  =  1. 

69.  Find  the  mean  value  of  log  x  from  x  =  1  to  a;  =  10. 

70.  Find  the  mean  value  of  logio  x  from  x  =  10  to  x  =  100. 

71.  A  body  moves  so  that  its  velocity  (in  feet  per  second)  is  propor- 
tional to  the  distance  travelled.  Express  the  distance  as  a  function  of 
the  time. 

72.  If  the  body  of  Example  71  travels  100  feet  in  10  seconds  and 
200  feet  in  15  seconds,  bow  far  will  it  travel  in  t  seconds? 

73.  The  acceleration  of  a  body  moving  in  a  straight  line  is  propor- 
tional to  the  distance  travelled.  If  the  velocity  is  zero  when  the  dis- 
tance is  zero  express  the  distance,  s,  as  a  function  of  t,  the  time. 


(d^s       dv     ds  dv\ 

Suggestion: -=-.-=.-j. 


LOGARITHMIC  FUNCTIONS  151 

74.  The  acceleration  of  a  body  moving  in  a  straight  line  is  inversely 
proportional  to  the  distance  travelled.  The  acceleration  is  1  ft.  sec.* 
when  the  distance  is  1  ft.  What  is  the  velocity  of  the  body  when  it 
has  travelled  10  feet,  H  v  =  0  when  5  =  1? 

75.  A  centre  of  force  at  a  point,  0,  attracts  a  particle,  which  is  at 
rest  a  feet  from  0,  so  that  the  acceleration  of  the  particle  is  equal  at 
any  instant  to  its  distance  from  0.  Find  the  velocity  of  the  particle 
when  it  has  travelled  halfway  to  0.    (See  suggestion  under  Example  73.) 


ANSWERS 

CHAPTER  I 
Art.  1 
1.  8  cu.  ft.;  343  cu.  ft.;  29.8  cu.  cm. 

6  lu  ^' 

fcc     ,       50/c     ,  k 

4.  7  =  — ;  I  =  — r~;  the  same,  -— - 
100  d^  200 

6.  3;  24;  21  6.  3;  6;  3  7.  2^;  8|;  5| 

8.  0;  0.6021  9.  1.5574;  —  0.8637;  —  2.4211 

10.  2;  16;  14  11.  —  2;  —  44;  —  42  12.  0;  —  60 

13.  —  4.8;  —  19.75;  —  14.95  14.  0;  —  1? 

Art.  3 

17.  Starts  from  (0,  —  2);  at  end  10th  sec.  (20,  8) 

18.  (3,3)  19.   (3.8,  —0.2)  20.  (1,  i) 
21.  (db  2.4,  =t  2.4)                                22.  (2.28,3.28);  (—3.28,-2.28) 
23.  (0,  0),  (=*=  .5,  =t  .5)                         24.  (.73,  2)  etc. 

25.  (0,  1),  (6.28,  1)  etc.  26.  (1.05,  1.73)  etc. 

27.  4^  sec,  8^  ft.  28.  3.5  +  sec,  7.5  +  ft. 

29.  2.9  sec;  6.9  ft.  and  3.4  +  ft.         30.  4.7  sec;  8.7  ft.  and  17.4  ft. 


CHAPTER  n 

Art. 

6 

1.  4  a: 

-!  +  ■ 

3.  6x  — 2 

4.  3^2 

6.  3t;2- 

1 

1)2 

6.2^  +  ^ 

7    Hmit    ^y 
limit    ^^ 

1 

1 

(y-iy 

153 


154  CALCULUS  AND  GRAPHS 


8. 

limit  ^y     1 

6 

X' 

9. 

limit 
Ax  = 

A2/           1 

0  A  X       2  V  x' 

limit 

A2/  = 

Ax       ^ 

o^y  =  '' 

10. 

limit   ^y  __ 

3 

limit    ^ 

3 

^x  =  0  ^j. 

(X- 

2)2' 

(2/ -1)2 

11. 

limit  ^y 

4 

limit    ^ 
A2/  =  0^2/ 

4 

Ax  =  OAa; 

(3  + 

^' 

(2/  +  1)2 

12. 

Hmit    ^V 

a 

limit    ^^ 
Art.  10 

=  — 

"p2 

1. 

9x2  —  2 

2. 

2x 

—  4  x3  +  3 

3.  3  (x  +  3)2 

4. 

2 

5. 

6x 

(X2    +  2)2 

6.  —  6  x2  (3  —  x3) 

(2  +  xY 

7. 

3rc2  +  l 
3  (0^3  +  xfl^ 

8. 

] 

—  4x 

9.  cos  2  X 

2  Vx-2a:2 

10. 

sin  a;  (1  +  sec^  x) 

11.  2  sec  2  X 

tan  2  X  —  sin  X 

12. 

2  cos  2  a:  —  3  sec^  3  x 

13.        ^/"^ 

(2^2- 

+  4) 
-4)2 

14. 

s2  —  28  s  —  93 

15. 

5  (3  2/2  +  2) 
2/2  (2/2  +  2y 

^^^  3(4  2/3-1) 
2/2(1-2/^)2 

s2  (s  +  7)2 

17. 

8.1  :cl-7 

18. 

25.1 

^6.02 

19.  1.57x2-14 

20. 

9 

2a:^^/^ 

21. 

dx            y 

22.  J^_if 
ax           5  2/ 

23. 

cZi/       4  a; 
dx       5y 

1 

24. 

^  _  __y 

dx            x 

25    di/_l+2/ 
dx        2  — X 

26. 

dy       3  —  2/ 

(io;  ~  a;  —  1 

27.  2  sin  (4  x 

+  ^l) 

28.  —  3  cot2  (x  —  I)  csc2  (x  —  I) 

29.  sec  (2  X  +  f )  {  2  cos  x  tan  (2  x  +  f )  —  sin  x  [ 

30.  cos  X  cos  2  X  tan  3  x  —  2  sin  x  sin  2  x  tan  3  x  +  3  sin  x  cos  2  x 

sec2  3  X 


ANSWERS 


155 


CHAPTER  m 

Art. 

13 

2. 

V  =  Sl^  —  Qth.  sec. 
/  =  6  ^  —  6  ft.  sec.2 
Starts  at  s  =  2  ft. 
Moves  S.  2  sec;  2  ft. 
Thereafter,  N, 

4. 

V  =  4  —  2  i  ft.  sec. 
/  =  —  2  ft.  sec.2 
Starts  at  s  =  5  ft. 

^ft. 

Moves  AT.  TF.  2  sec;  9  ft. 
Thereafter,  S.  E. 

6. 

V  =  ^t  —  Zf^it.  sec. 
/  =  6  —  6  Ht.  sec.2 
Starts  at  s  =  0 
Moves  N.  2  sec;  4  ft. 
Thereafter,  S, 

1.  i;  =  6  i  —  3  <2  ft.  sec. 
/  =  6  —  6  i  ft.  sec.2 
Starts  at  s  =  2  ft. 
Moves  ^.  2  sec;  6  ft. 
Thereafter,  W. 

3.  V  =  18  i  —  5  ft.  sec. 
/  =  18  ft.  sec2 
Starts  at  s  =  7  ft. 
Moves  S.  W.  xV  sec;  6^  ft, 
Thereafter,  N.  E. 

6.  V  =  St^  —  Qth.  sec. 
/  =  6  <  —  6  ft.  sec2 
Starts  at  s  =  0 
Moves  W.  2  sec ;  4  ft, 
Thereafter,  E 

V  =  32  t,  64,  160  ft.  sec;/  =  32  ft.  sec^ 

r  =  32  i  +  3,  99,  323  ft.  sec;  /  =  32  ft.  sec* 

320  ft.  sec  10.  320.8  ft.  sec. 

10  sec,  —  1600  ft.  12.  6^  sec,  625  ft. 

143.1  ft.  sec. 

64  ft.  sec;  6.4  ft.;  6.56  ft.;  .16  ft.  =  2.4%. 

.64  ft.;  .6416  ft.;  .0016  ft.  =  .25%. 


7. 

8. 

9. 
11. 
13. 
14. 

16. 
17. 
19. 
20. 
23. 

26. 
26. 
27. 


■g-  in.  per  min. 
5  in.  per  min. 

V  =  2.65;  10.22- 

67  2^ 
~10^'    10^ 


14  t;  ■ 


16.  .069  in.  per  sec. 
18.  1  cm.  per  sec;  faster. 
17.78;  neither  — why? 

21.  1.009  22.  0.172 


305  rad.  sec, 
120  rad.  sec^ 


24.  200  X  —  40  X  /,  —  40  x, 
1  rev.  per  sec,  5  min. 

1000  X  VS  in.  per  mm., — 1000  x  in.  per  min. 

—  4.2,  2.7,  i  =  1;  0,  —  5,  ^  =  2  X. 

9,  W%  t  V2.  28.  Former  by  0.024  sec. 


156 


CALCULUS  AND  GRAPHS 


CHAPTER  IV 
Art.  14 

-  6  x,  7,  6.7 

1 
5.  — - 


1.  4  X,  8,  8.2  2. 

4.  3a;2  — 2  a:,  21,  21.81 

40 
6.  -",-5,-4.65 

x^ 

8.  6  i  —  3  i^  —  9,  —  9.61 

9.  Greater  when  ^  =  2;  t;  =  8 

10.  Greater  when  t  =  2;  v  =  3f 

11.  Greater  when  t  =  1;  v  =  12 

12.  Greater  when  t  =  2;  v  =  12 

13.  Greater  when  t  =  2;  v  =  12 

14.  Greater  when  t  =  2;  v  =  S 

15.  —  2,  0,  1  16.  0,  —  2,  V3 
18.  i,  1  19.  0,  00 


3.  3  a;2  +  2,  14,  14.61 
5 


2' 


-2.38 


7.  3  ^2  __  6  ^^  9^  9  61 


17.  1,  2 

20.    oo,-fV3 


Max.  |2 
Min.  0 


6.  Max.  16 


Art.  15 
2.  Max.  ^J_ 
Min.  0 


Min.  —  iJl 


7.  Max.  1 
Min.  —  1 


3.  Neither 

4.  Neither 

5.  Neither 
Max.  \/2 
Min.  —  V2 


12.  Min.  14f  ft.  sec. 


8.  Max.  V2 
Min.  —  \/2 

10.  Min.  2f  ft.  sec.  11.  Max.  1^  ft.  sec. 

13.  2^  ft.,  2i  ft.  14.  108.6  cu.  in. 

15.  Height  equals  diameter  of  base. 

16.  15406  +  sq.  yds.  17.  Height  =  width  of  sill  =  4.2  ft 

u  sin  a  u^  sin^  a 

18.  t  =  ,  X  =  — 

g  ^9 

19.  Height,  4  V3;  diameter,  4  \/6;  volume,  96  t:  VS 

20.  Height,  8.6  ft.;  diameter,  12.2  ft.;  volume,  331  cu.  ft. 


ANSWERS  157 

21.  2  X  (1  —  \/f)  ==  "g"»  approximately. 

22.  Height,  11.5  ft.;  surface,  52.6  sq.  ft. 

««    16x 

23.  ——z  cu.  ft. 
9V3 

24.  a  =  900,      6  =  60000,      m  =  3000 «  =  -^""  vJU 


-(i^r 


^i-r     ^4-i' 


25.  -  26.  -  27.  3  hrs.  40  min. 

4  2 

28.  No;  nearest,  3^  miles,  after  25  minutes. 

29.  Stem,  12  ft.  6  in.;  arm,  6  ft.  11  in.,  approx. 

30.  Length  along  boundary  =  1^  times  width. 

31.  Width  (side  of  triangle),  8.2  ft.;  height  (of  rectangle),  5.2  ft. 

32.  Diameter, :  side, 

'    X  +  1'        '2(x  +  l) 

«-.    ,.    ,.  3/30  ,    .  ,      20  3/30", 

33.  Radius,  5  \/— ;  height,  —  \/—  ft. 

V    X  3   V    X 

34.  6  ft.  10  in.,  approx. 

Art.  16 


1. 

3  X  (x  —  2)  rfx 

2. 

(1  +  2  X  —  4  x3)  dx 

3.  (3x2— |a;V=)da 

4. 

(3x2  4.  1)  ^:p 
2  Vx^  +  X 

5. 

2xdx 

4xdx 

3  (x2  —  2)V» 

''•  3  (x^  +  1)V. 

7. 

3  sin  6  X  dx 

8. 

5  sec^  5xdx 

X            X 

9.  sec2-tan-dx 
2        2 

10. 

(1  +  cos  x)  dx 

11.  —  3  esc2 

(3  X  +  f )  dx 

12. 

—  2sin(2x  +  |) 

dx 

13.  -^dx 

X 

14.  -^.ix 

X 

15. 

2x 

16.   'i^:/,. 

17.  —-dx 

y 

158  CALCULUS  AND  GRAPHS 

18.  —  dx  19.  115.2  X  cu.  ft. 

32/ 

25  X    9  X 
20.  197  ft.,  9.8  ft.  21.  -— ,  —-  cu.  ft. 

4         4 

22.  dv  =    izr'^  cot  a  dr 

23.  14.8  X  sq.  in.,  136.9   x  cu.  in. 

24.  0.16  sq.  in.  25.  9.6  cu.  in.  26.  24.24 

27.  In  each  case  approximate  value  is  unity. 

Exact  value  (1)  1.00007525;  (2)  1.00007475 

28.  .006 

29.  (a)  83i  sq.  ft.;  (b)  62^  sq.  ft.;  (c)  —  145|  sq.  ft. 

30.  (a)  5  ft.;  (b)  —  5  ft.;  (c)  —  1.4  ft. 

31.  Errors  in  C  and  E  both  =t,  =f  .005;  errors  in  C  and  E  one  =»=,  one  =f 

=T=  .055;  about  .9%. 

32.  .1015  33.  .005  ft.  sec.2 
34.  .018;iofl%.  35.  .03;  4%. 

36.  —0.025 

37.  (a)  =*=  .008;  (b)  =f  .016;  (c)  —  .008;  (d)  .024 

CHAPTER  V 

Art.  18 

Lx+ix^^  2.  fa:V3  +  J  3.  |-2V^ 

4.  f  x'l'  {x'h  —  2)  ^-  f  +  *  ^'''  ^-  i  ^'^'  +  i  ^'^' 

13.  a;  +  I  xl^  +  -  14.  a:  —  3  x^l'  +  3  xl' 

2         1  z^  3 

15.  a:  + -—  16.  -  +  2  0  —    -; 

X       Zx^  5  2^ 

17.  1 2/^/3  -  2  2/  +  3  2/V3  18.  |  +  H  ^"^^  +  I  ^'/» 


ANSWERS  159 

1ft     ^    n-^f^x)-^  20.  — -^ 21.  —  I  (4  —  3  O'' ' 

19.  ^TT  (^3  —  5  X)  ^u.       2  (1  +  2  x)  ' 

22.  i  (2  i/  +  1)'/^  23.  ^  VTTT0  24.  -  f  V2-3« 

25.   1  (0:2  +  1)4  26.  —  ^  (2  -  t^)'  27.  ^  (S  +  u^V 

oo I 29.  ' 30.  —  tV(2  — a:^)' 

^®-        12(3+2x^)2  ^^-8(5-5^)2 

31.  I  \/27m^  32.  -  V-  ^1^^/^  33.  I  (2  x«/=  +  1)V* 

34.  I(x2  +  2  x)'/-^  35.  V  :c2  +  2  X  36.  Ij  (3  x^  —  6  x)*/'« 

37.  f  sin3  -  38.  —  i  esc  2u  39.  —  2  -y^  cos  < 
2 

40.  —  ^2  cos^  3  X  41.  I  sec2  3  ^^  42.  i  sin^  4  x 

X2                                                                                  a;2  X*     ,      - 

AS.y=ix^--  +  5  44.  i/  =  ---  +  T 

45.  2/  =  |-  +  ^"— ^/                     46.  y  =  -  +-  +2X-V- 


5^2 


il+4i--V-  48.  2/=4i-—  +6 

2  ^ 


47.  2/ 

49.  2/  =  16i2_37i4.26  60.  2/  =  16i2_56/+44 

51.  2/  =  2  x2  +  3  X  —  3  52.  2/  =  3^2  —  5  X  +  8 

53.  6  =  16i-2i2_io  54.  s=62  +  16«  +  Y 

55.  .  =  16^+  — -"ff  66.  s  =  16«-|<2_^« 

57.  s  =  7.1  ^2  +  c;  28.4  ft.,  21.3  ft.,  35.5  ft. 

3  x2 

58.  2/  =  3x  +  c  59.  y  =  — — 2x  +  c 

60.  y  =  3x  — ^— ^  +c  61.  2/  =  2^'+c 

62.  2/  =  -  —  0:2  +  3  X  +  c  63.  2/  =  x^  +  c 


160  CALCULUS  AND  GRAPHS 


64.  s  =  4  ^  +  c 

65.  s  ==t^  +  St  +  c 

66.  s  =  ^3  +  ^2  _ 

3^-1- 

■  c                      67.  s  =  4  J 

'-I- 

68.  s  =  t^  +  c 

69.  s  =  — 

CHAPTER  VI 
Art.  19 

7- 

1.  15.12;  15.02 

2.  15.25;  15.32 

3.  10.5;  10.64 

4.  35.75;  35.96 

5.  1.81;  1.83 

6.  2.62;  2.66 

7.  1.69;  1.702 

8.  3.97;  3.996 

9.  41;  40.16 

10.  61.5;  60.24 

11.  1.396 

12.  1.974 

13.  1.396 

14.  1.974 

15.  119.25  sq.ft. 

16.  1410  sq.ft. 

17.  1260  sq.  ft. 

18.  5.32 

19.  42.5 

20.  1400  sc 

CHAPTER  Vn 
Art.  20 

I.  ft. 

1.  1                  2. 

63 

3.  ^51| 

4.  2                5.  —i 

6.  ^i             7. 

1 

20 

8.  7-^ 
Art.  22 

9    2^2-^ 
3 

1.  12f                2, 

.  —6 

f                3.  lOf 

4.  36                  5.  If 

6.  2f                 7. 

.  iM 

8.  4 

9.  40                 10.  60 

11.  V2 

12. 

2                       13.  V2 

14.  0  or  2 

15.  1280  sq.ft.;  1422f  J 

3q.  ft. 

16.  5i 

17. 

V2  —  1                 18.  2  - 

-  V2                    19.  i 

20.  V2(l— f) 

21. 

1  +  V_2    (^_4) 
Art.  23 

22.  85i        23.  i 

136  X 
^'       3 

2. 

40  X                         5312 
3                         '       15 

X                    ^    235  X 
*•       6 

ANSWERS 

161 

^    344x 
^-      3 

6. 

8_x 
3 

(7  +  6  V^) 

7.  f 

X  ab^;  ^  X  a% 

8.  f  X  a»           9. 

75 

t: 

-TT 

15 

12.  36  X 

13.  y(2-V2) 

14.  - 

f       -» 

X            16.    x(l  — f) 

: 

"•¥ 

-f 

Art.  24 

1.  3fl  tons 

2. 

7Htons 

3. 

6.6  tons 

4.  13.2  tons 

6. 

ft  tons 

6. 

1-^^  tons 

7.  151.6  lbs. 

8. 

19^1-  tons 

9. 

500  lbs.;  1^  ft. 

10.  8.4  in. 

11. 

aW2^ 
^     tons 
64 

12. 

YS  tons 

13.              tons 
3 

14. 

3V3^ 

tons 
5 

15. 

Side,  l^  tons; 
end  f  tons 

16.  951561^  tons 

Art.  25 

,    ,  .  625V3       ^  „,  1875  V3..    ^ 

1*  (a)   """r; —  it.  tons  (b)  — — —  ft.  tons 

64  64 

(c)  38  ft.  tons,  nearly  (d)  134.9  ft.  tons 

(e)   Iff  ft.  tons  (f)   3f|-ft.  tons 

2.  27f  f  ft.  tons        3.  30|4  ft.  tons        4.  i  ft.  ton 

Art.  26 


1.  (a)  112.5  ft.; 

(b)71 

5  ft. 

2.  61  ft. 

3.  5131  ft. 

^    4500 

5.            revs. 

X 

684 
6.  —  revs. 

X 

7.  f  sec.;  2  ft. 

8.  4  ft. 

9.1 

10.  50  ft.;  5  sec. 

11.  3Ht.;l^sec. 

12.  3|ft.;l^sec. 

13.  4  ft.;  4  sec. 

14.  5^  ft.;  4  sec. 

16.  4i  ft.;  3  sec. 

162        CALCULUS  AND  GRAPHS 

16.  170f  ft.;  16  sec.  17.  l|  rads.;  f  sec;  2^  ft. 

18.  I  rad.;  ^  sec;  1  ft.  19.  2^  rad.;  1^  sec;  4^  ft. 

X^     X     x^ 
20.  -ox  rad.;  i  sec;  1  in.         21.  —  rad.;  —  sec;  —  ft. 
2  4.  J  2         ,  24     2     12 

X^     X     x^ 
22.  —  rad.;  -  sec;  —  ft.         23.  10  sec;  250  ft. 
96     2    '  48 

24.  llf  sec;  5^  ft.  sec^  25.  250  ft. 

26.  ^  sec;  .15  sec;  .35  sec;  700  ft.  sec 

Art.  27 

1.  (a)  ^^  ft.     (b)  ^^7^  ft.      (c)  7.7  +  ft.  below  surface. 

2  4 

(d)  10.2  ft.  (e)  2ift.  (f)  3fft. 

(g)  1.4  ft.  below  top  of  gate.  (h)  2  ft. 

2.  (4,  li)  *  3.  di  1)  *  4.   (1,  li)  * 
5.  (2tV,  I)  *  or  (2H,  I)  6.   (2f ,  0)    _  7.   (0,  2f ) 

11.   a,  I)  12.  (f ,  i)  13.   (I,  4) 


Art.  28 

1.  3|;  4i  2.    I  3.  3  4.  4  6.  7 

6.  2f;2i  7.  2;  5  8.  51;  153;  153 

9.  66.1  nearly  10.  61.5  lbs.;  62  lbs.  11.  2  or  1^ 

2  a2  2  X  a2  X  r2 

12.  3;  3  13.  — -  14.    -——  15.  —- 

3  3  3 

16.  16  ft.  sec;  96  ft.  sec.  17.  33.6  lbs.  per  sq.  ft. 

18.  -  19.  6i;  3i  20.  0.7911 

2 

21.  i  22.  i  23.  0 

*  One  vertex  at  the  origin. 


ANSWERS  163 

Art.  29 

1.  7333^  ergs.  2.  t|^  dyne.  3.     f^     . 

«  (fc  +  «) 

4.  6.  3.79  foot-tons  8.  

2  3 

9.^  ,0.,.,'^K;^,'^K 


CHAPTER  Vra 
Art.  30 

3.  144.8  max.;  — 10.03  min. 

4.  Axis  of  X  at  tan~^  99,  tan"^  ( —  18),  tan"^  22;  axis  of  y  at  tan"*( —  34). 

5.  Axis  of  X,  198,  —  36,  44  ft.  sec;  axis  of  y^  —  68  ft.  sec;  max.  and 

min.  pts.,  0. 

6.  87f;— 6t;70.6or73. 

7.  Yes;  at  x  =  —  3.8  and  x  =  3.1  8.  803.1 
9.  —  3.1                              10.  3.02                              11.  —  3.38  or  0.63 

12.  104;  1.72  16.  (a)  2.5  tons;  (b)  2.46  tons;  (c)  2.48  tons 

17.  2.48  tons  18.  2.48  tons  19.  8 

20.  2500  units  per  min.  21,  0.1  ft. 


_    (fe  — 2)    =t=   V(&  — 2)2  —  IQq 

a 

25.  Yes;  1  sec;  7  ft. 

26.  1  sec;  7  ft.;  or,  (Query  ?)  if  sec;  8|  ft. 

27.  (a)  V4l;  (b)  lA;  ^^;  (c)  lA 

X   3x 

28.  Always  same  distance  apart;  ^  =  -,  — ,  etc. 

4      4 

29.  Nearest;  ^  =  0,  x,  2  x  etc.;  0 

^     ,  X   3x 

Farthest;  t  -  -,  —  etc.:  2 


164  CALCULUS  AND  GRAPHS 

2  X 
30.  —  —  2  31..  8  sec;  365.3  ft.;  67  ft.  sec. 

Z4:.  I  =  a  Z5.  R  =  -  36.  0;  —  13.2  ft.  min. 

37.  20  min.  55  sec;  2  min.  13  sec 


38. 


39.  -VgR;     \/2  gR  40.  4.9  and  7  miles  per  sec 

64  X 
42.  17t^5  sq.  ft.  43.  -—  cu.  ft.  44.  tan"!  (—  2) 

o 

45.  f ;  tan-l  (—  2)  46.  4,  4,  -^^  ft.  sec 

o 

47.  ^  =  0.9  sec;  (0.8,  0.7)  48.  - 

o 

49.  100  ft.;  40  ft.  sec.  50.  53^^  ft.  sec. 

53.  At  a;  =  1,  each  is  parallel  to  OX;  at  x  =  1.1,  32°  13',  5°  22'  and 

0°  43'. 

54.  6.75;  —  12.15;  24.3 

55.  At  re  =  1  each  is  infinite 

At  X  =  1.1,  3.75  ft.  sec,  21.37  ft.  sec,  160.02  ft.  sec. 

56.  2  57.  12^  oz.  approx.  58.  2;  12 

2  X 
63.  (a)  Max.  2.598  at  x  =  —  etc.; 
o 

10  X 
Min.  —  2.598  at  x  = etc 

o 

(b)  Max.  0.369  at  a:  =  122°  32' 

1.760  at  a:  =  323°  37' 

Min.  —  0.369  at  x  =  57°  28' 
—  1.760  at  x  =  216°  23' 

(c)  Max.  0.607  at  re  =  117°  26' 

Min.  —  0.607  at  a;  =  242°  34' 


ANSWERS  165 

(d)  Max.  f  at  x  =  0 

4  X 
0.216  at  X  =  —- 
5 

2  X 
Min.  —  0.674  at  a;  =  — - 
5 

^  at  X  =   X 
64.  (a)8;(b)-i;(c)l 


CHAPTER  rX 
Art.  35 

1.  None  2.  None  3.  Min., at  a;  =  —  1 

e 

4.  Max.,  -  at  X  =  1  6.  Min.,  e  at  x  =  1 

e 

4 

6.  Min.,  0  at  X  =  0;'  Max.  —  at  x  =  2 

7.  None  8.  Min.,  1  at  x  =  0  9.  None 

10.  ^1  =  —  — ^— ;      A2  = ^^ 

2^^  2e^ 


ji+.-^);A.  =  -f{^-f^^..) 


11.  .4i  =  |  |l+e  2|;  ^2  = 

12.  X  =  -,    X,  — ,  3  X  etc.  13.  x  =  0,  -,  2  x,  —  etc. 

2'      '    2  '  '  2'        '   2 

y  =  Qy.?  y  =  Qy.? 

14.  X  =  —  .33  ft.  16.  X  =  .996  ft. 

v  =  —  .99  ft.  sec.  v  =  —  2.54  ft.  sec. 

/  =  1.53  ft.  sec.2  /  =  3.1  ft.  sec.2 

t;  =  0  when  i  =  .76  sec,  ^     ,  3  x    7  x 

r>oA    r^  r.4^^  V  =  0  when  i  =  — — ,  — 7- 

2.34  sec.  etc.  4  '     4 

etc.  sec. 

Max.  t;  =  1.45  ft.  sec.  _^  5   . 

Max.  y  =  -i:  ft.  sec;  Max.  v 

(numerical)  5  ft.  sec 


166  CALCULUS  AND  GRAPHS 


16.  a;  =  2e«ft.  17.  x  =  2e  ^  ft. 

X  X 

V  =  e^  ft.  sec.  V  =  1.46  e     ft.  sec. 

X  _J7C 


/  =  —  -#  e"*  ft.  sec.2  /  =  —  3.46  e  ^  ft.  sec. 

4'  "i 


V  =  0  when  i  =  .9,  2.5  etc.  sec.  ?;  =  0  when  <  =  — ,  etc. 


sec.  ^  3j£ 

Max.  iJ  =  4.5  ft.  sec.  Max.  v  =  4:6     ^  ;  Max.  v  (nu- 

X 

merical)  4  e 


18. 

a;2— 1 

^S-.^ 

22. 

2a;  +  1 

23.     ^^' 

x^  +2x 

x2+a; 

25. 

Ai  =  8.82; 

^2 

=  9.68 

27. 

52. 

28.   -4^— 

20.  21.  

X  —  x^  x"^  ■\-  X 

24.  8.05 

2x 

26.  , — - 

log  2 

29.  ie2^+l 


2  log  5  4  log  X 

30.  4  a;  31.  log  sec  x  32.  log  \/2  x  +  1 

33.  log  {x  —  5)  34.  \  log  (3  a;  +  2)  35.  log  {^  +  1) 

36.  i  log  (2  e2^  —  5)  37.  log  (log  x)  38.  log  (1  +  sin  x) 

39.  log  V2  tan  x  +  3     40.  —  ^ log  (cos 2  a;  +  4)     41.  (^~"~) 

(1  \  e^x 

—  +  1  I  43.  -7-  (2  X  —  1)  44.  sin  x  —  a;  cos  x 

64  /  4 

45.  |-  (2  a;  sin  2  a:  +  cos  2  x)  46.  2  cos  a;  +  2  a;  sin  a;  —  x^  cos  a; 

47.  a;  sin-l  ^  _|.  ^^  _  ^2  43,  3,  ^^n"!  a;  —  log  y/l  +  a;2 

49.  —  cos  e^  50.  i  e^'  51.  

e  +  1 


52.  f  (e*— 1)  53.  f  54.  x 

55.  f  (e2  +  1) 

56.  2;  =  a  {e^  —  e"0 ;  /  =  a  (e'  +  ^"0  =  s;  no  minimum 


(•— ji) 


ANSWERS  107 

57.  s  =  a  (e5  +  e"^  —  2);  t;  =  a  (e^  —  g-S) 

58.  .00134  59.  .0011  60.  .1181 

61.  N  =  CE^  62.  -^  log  ^'  63.  p  =  ce"^'^ 

64.  762  e'^^;  30  e"**  65.  X  =  -^^;  4  min.  42  sec.;  8  min.  2  sec. 

200  ' 


66. 

K  =  log2 

67.  3  min.  19  sec. 

68.  1.718 

69. 

1.5584 

70.  1.6768 

71.  S  =  CE^ 

72. 

t 
25(2)-' 

73.  S  ^CE^"^^ 

74.  2.15  ft.  sec. 

75. 

--:v3 

:  ^   N 


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